Question on Induction [closed]

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The question I have has to do with a question we were given in class for homework. The question is:

For all integers n>=1, 1+6+11+16+...+(5n-4) = (n(5n-3))/2

I got the left hand side on my own, it's true. I found the solution to the right hand side, which was:

=$$(k(5k-3))/2 + (5(k+1)-4)$$
=$$(5k^2-3k+10k+10-8)/2$$
=$$(5k^2+7k+2)/2$$
=$$((k+1)(5k+2))/2$$
=$$(k+1)(5(k+1)-3)/2$$
=$$right hand side, thus P(k) is true. Therefore by principles of mathematical induction P(n) is true for all integers.$$

This is the step I don't understand. I'm not following anything they got after the 3k.
I'll leave it like that so if someone's trying to figure it out, I don't screw it up. So, if someone doesn't mind explaining that, I'd appreciate it.

induction

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edited Jul 20 at 0:40

asked Jul 20 at 0:23

Wantopinions

11

closed as unclear what you're asking by Shailesh, Arnaud Mortier, user223391, Xander Henderson, Isaac Browne Jul 20 at 2:07

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  What do you mean by the left hand side being "true"? What do you mean by the right hand side having a "solution"? What does $k$ have to do with anything? Please show the work you did, so we can figure out what's going wrong, if anything.
  – Cameron Buie
  Jul 20 at 0:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Write it as $,1 + 5 cdot big(1 + 2 + ldots+(n-1)big),$. What's inside those parentheses should look familiar.
  – dxiv
  Jul 20 at 0:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I'm not sure what you mean by 'true.' If you want to prove this by induction, you begin by establishing the base case (prove for $n = 1$ that the equality holds), and then prove that it holds for $n = k + 1$, given that it holds for $n = k$. If you'd like a more thorough explanation of that, let me know and I'll write up a longer answer for this. But I think there might be a misunderstanding on either my end in reading the question or perhaps on the instructions, as what you've written doesn't seem to be a proof by induction.
  – Matt.P
  Jul 20 at 0:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Wantopinions Is that the induction step in a tentative proof by induction, by any chance? Then write so in the question, with all the details. As it stands now it is incomprehensible what the question actually is, where $n$ disappeared, who are they etc.
  – dxiv
  Jul 20 at 0:39
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Your syntax doesn't make any sense right now, as indicated by previous comments. I'm voting to close the question until it is fixed.
  – Arnaud Mortier
  Jul 20 at 0:45
  

  
  
  
  

add a comment | 

up vote
-3
down vote

favorite

The question I have has to do with a question we were given in class for homework. The question is:

For all integers n>=1, 1+6+11+16+...+(5n-4) = (n(5n-3))/2

I got the left hand side on my own, it's true. I found the solution to the right hand side, which was:

=$$(k(5k-3))/2 + (5(k+1)-4)$$
=$$(5k^2-3k+10k+10-8)/2$$
=$$(5k^2+7k+2)/2$$
=$$((k+1)(5k+2))/2$$
=$$(k+1)(5(k+1)-3)/2$$
=$$right hand side, thus P(k) is true. Therefore by principles of mathematical induction P(n) is true for all integers.$$

This is the step I don't understand. I'm not following anything they got after the 3k.
I'll leave it like that so if someone's trying to figure it out, I don't screw it up. So, if someone doesn't mind explaining that, I'd appreciate it.

induction

share|cite|improve this question

edited Jul 20 at 0:40

asked Jul 20 at 0:23

Wantopinions

11

closed as unclear what you're asking by Shailesh, Arnaud Mortier, user223391, Xander Henderson, Isaac Browne Jul 20 at 2:07

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  What do you mean by the left hand side being "true"? What do you mean by the right hand side having a "solution"? What does $k$ have to do with anything? Please show the work you did, so we can figure out what's going wrong, if anything.
  – Cameron Buie
  Jul 20 at 0:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Write it as $,1 + 5 cdot big(1 + 2 + ldots+(n-1)big),$. What's inside those parentheses should look familiar.
  – dxiv
  Jul 20 at 0:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I'm not sure what you mean by 'true.' If you want to prove this by induction, you begin by establishing the base case (prove for $n = 1$ that the equality holds), and then prove that it holds for $n = k + 1$, given that it holds for $n = k$. If you'd like a more thorough explanation of that, let me know and I'll write up a longer answer for this. But I think there might be a misunderstanding on either my end in reading the question or perhaps on the instructions, as what you've written doesn't seem to be a proof by induction.
  – Matt.P
  Jul 20 at 0:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Wantopinions Is that the induction step in a tentative proof by induction, by any chance? Then write so in the question, with all the details. As it stands now it is incomprehensible what the question actually is, where $n$ disappeared, who are they etc.
  – dxiv
  Jul 20 at 0:39
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Your syntax doesn't make any sense right now, as indicated by previous comments. I'm voting to close the question until it is fixed.
  – Arnaud Mortier
  Jul 20 at 0:45
  

  
  
  
  

add a comment | 

up vote
-3
down vote

favorite

up vote
-3
down vote

favorite

The question I have has to do with a question we were given in class for homework. The question is:

For all integers n>=1, 1+6+11+16+...+(5n-4) = (n(5n-3))/2

I got the left hand side on my own, it's true. I found the solution to the right hand side, which was:

=$$(k(5k-3))/2 + (5(k+1)-4)$$
=$$(5k^2-3k+10k+10-8)/2$$
=$$(5k^2+7k+2)/2$$
=$$((k+1)(5k+2))/2$$
=$$(k+1)(5(k+1)-3)/2$$
=$$right hand side, thus P(k) is true. Therefore by principles of mathematical induction P(n) is true for all integers.$$

This is the step I don't understand. I'm not following anything they got after the 3k.
I'll leave it like that so if someone's trying to figure it out, I don't screw it up. So, if someone doesn't mind explaining that, I'd appreciate it.

induction

share|cite|improve this question

edited Jul 20 at 0:40

asked Jul 20 at 0:23

Wantopinions

11

The question I have has to do with a question we were given in class for homework. The question is:

For all integers n>=1, 1+6+11+16+...+(5n-4) = (n(5n-3))/2

I got the left hand side on my own, it's true. I found the solution to the right hand side, which was:

=$$(k(5k-3))/2 + (5(k+1)-4)$$
=$$(5k^2-3k+10k+10-8)/2$$
=$$(5k^2+7k+2)/2$$
=$$((k+1)(5k+2))/2$$
=$$(k+1)(5(k+1)-3)/2$$
=$$right hand side, thus P(k) is true. Therefore by principles of mathematical induction P(n) is true for all integers.$$

This is the step I don't understand. I'm not following anything they got after the 3k.
I'll leave it like that so if someone's trying to figure it out, I don't screw it up. So, if someone doesn't mind explaining that, I'd appreciate it.

induction

share|cite|improve this question

edited Jul 20 at 0:40

asked Jul 20 at 0:23

Wantopinions

11

share|cite|improve this question

share|cite|improve this question

edited Jul 20 at 0:40

edited Jul 20 at 0:40

edited Jul 20 at 0:40

asked Jul 20 at 0:23

Wantopinions

11

asked Jul 20 at 0:23

Wantopinions

11

asked Jul 20 at 0:23

Wantopinions

11

11

closed as unclear what you're asking by Shailesh, Arnaud Mortier, user223391, Xander Henderson, Isaac Browne Jul 20 at 2:07

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

closed as unclear what you're asking by Shailesh, Arnaud Mortier, user223391, Xander Henderson, Isaac Browne Jul 20 at 2:07

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  What do you mean by the left hand side being "true"? What do you mean by the right hand side having a "solution"? What does $k$ have to do with anything? Please show the work you did, so we can figure out what's going wrong, if anything.
  – Cameron Buie
  Jul 20 at 0:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Write it as $,1 + 5 cdot big(1 + 2 + ldots+(n-1)big),$. What's inside those parentheses should look familiar.
  – dxiv
  Jul 20 at 0:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I'm not sure what you mean by 'true.' If you want to prove this by induction, you begin by establishing the base case (prove for $n = 1$ that the equality holds), and then prove that it holds for $n = k + 1$, given that it holds for $n = k$. If you'd like a more thorough explanation of that, let me know and I'll write up a longer answer for this. But I think there might be a misunderstanding on either my end in reading the question or perhaps on the instructions, as what you've written doesn't seem to be a proof by induction.
  – Matt.P
  Jul 20 at 0:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Wantopinions Is that the induction step in a tentative proof by induction, by any chance? Then write so in the question, with all the details. As it stands now it is incomprehensible what the question actually is, where $n$ disappeared, who are they etc.
  – dxiv
  Jul 20 at 0:39
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Your syntax doesn't make any sense right now, as indicated by previous comments. I'm voting to close the question until it is fixed.
  – Arnaud Mortier
  Jul 20 at 0:45
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  What do you mean by the left hand side being "true"? What do you mean by the right hand side having a "solution"? What does $k$ have to do with anything? Please show the work you did, so we can figure out what's going wrong, if anything.
  – Cameron Buie
  Jul 20 at 0:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Write it as $,1 + 5 cdot big(1 + 2 + ldots+(n-1)big),$. What's inside those parentheses should look familiar.
  – dxiv
  Jul 20 at 0:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I'm not sure what you mean by 'true.' If you want to prove this by induction, you begin by establishing the base case (prove for $n = 1$ that the equality holds), and then prove that it holds for $n = k + 1$, given that it holds for $n = k$. If you'd like a more thorough explanation of that, let me know and I'll write up a longer answer for this. But I think there might be a misunderstanding on either my end in reading the question or perhaps on the instructions, as what you've written doesn't seem to be a proof by induction.
  – Matt.P
  Jul 20 at 0:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Wantopinions Is that the induction step in a tentative proof by induction, by any chance? Then write so in the question, with all the details. As it stands now it is incomprehensible what the question actually is, where $n$ disappeared, who are they etc.
  – dxiv
  Jul 20 at 0:39
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Your syntax doesn't make any sense right now, as indicated by previous comments. I'm voting to close the question until it is fixed.
  – Arnaud Mortier
  Jul 20 at 0:45
  

  
  
  
  

2

2

What do you mean by the left hand side being "true"? What do you mean by the right hand side having a "solution"? What does $k$ have to do with anything? Please show the work you did, so we can figure out what's going wrong, if anything.
– Cameron Buie
Jul 20 at 0:26

What do you mean by the left hand side being "true"? What do you mean by the right hand side having a "solution"? What does $k$ have to do with anything? Please show the work you did, so we can figure out what's going wrong, if anything.
– Cameron Buie
Jul 20 at 0:26

Write it as $,1 + 5 cdot big(1 + 2 + ldots+(n-1)big),$. What's inside those parentheses should look familiar.
– dxiv
Jul 20 at 0:29

Write it as $,1 + 5 cdot big(1 + 2 + ldots+(n-1)big),$. What's inside those parentheses should look familiar.
– dxiv
Jul 20 at 0:29

I'm not sure what you mean by 'true.' If you want to prove this by induction, you begin by establishing the base case (prove for $n = 1$ that the equality holds), and then prove that it holds for $n = k + 1$, given that it holds for $n = k$. If you'd like a more thorough explanation of that, let me know and I'll write up a longer answer for this. But I think there might be a misunderstanding on either my end in reading the question or perhaps on the instructions, as what you've written doesn't seem to be a proof by induction.
– Matt.P
Jul 20 at 0:37

I'm not sure what you mean by 'true.' If you want to prove this by induction, you begin by establishing the base case (prove for $n = 1$ that the equality holds), and then prove that it holds for $n = k + 1$, given that it holds for $n = k$. If you'd like a more thorough explanation of that, let me know and I'll write up a longer answer for this. But I think there might be a misunderstanding on either my end in reading the question or perhaps on the instructions, as what you've written doesn't seem to be a proof by induction.
– Matt.P
Jul 20 at 0:37

@Wantopinions Is that the induction step in a tentative proof by induction, by any chance? Then write so in the question, with all the details. As it stands now it is incomprehensible what the question actually is, where $n$ disappeared, who are they etc.
– dxiv
Jul 20 at 0:39

@Wantopinions Is that the induction step in a tentative proof by induction, by any chance? Then write so in the question, with all the details. As it stands now it is incomprehensible what the question actually is, where $n$ disappeared, who are they etc.
– dxiv
Jul 20 at 0:39

Your syntax doesn't make any sense right now, as indicated by previous comments. I'm voting to close the question until it is fixed.
– Arnaud Mortier
Jul 20 at 0:45

Your syntax doesn't make any sense right now, as indicated by previous comments. I'm voting to close the question until it is fixed.
– Arnaud Mortier
Jul 20 at 0:45

add a comment | 

2 Answers
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up vote
1
down vote

I thought I would write out a answer that hopefully clarifies how you would approach this proof by induction.

The principle of induction states that, if we establish a 'base' case, usually $n = 0$ or $n = 1$ (depending on whether your class or textbook takes $0$ to be a natural or not, and depending on the particular statement you're trying to prove), and that it holds for some arbitrary natural number, $k$, then it holds for $k + 1$. Then, since you established $1$ as a basis, it holds for $2$, and since it holds for $2$, it holds for $3$, and so forth, until we cycle through all of the naturals. There are sometimes nuances in how to approach the base case, but this is the most general approach and all you need in this particular problem.

So, your goal is to prove that, for $n geq 1$, $sumlimits_i=1^n 5i - 4 = fracn(5n - 3)2$.

Let's start with the base case, which is $n = 1$ in this case. The left-hand side evaluates to $5(1) - 4 = 1$, while the right-hand side evaluates to $frac1(5(1) - 3)2 = 1$. So, this checks out.

Next, we assume that this statement is true for $n = k$, our 'induction hypothesis.' In other words, assume that
beginalign*
sumlimits_i=1^k 5i - 4 = frack(5k-3)2.
endalign*
We want to establish that the statement holds for $n = k + 1$. With a little bit of algebra, we can see that our goal is
beginalign*
sumlimits_i=1^k+1 5i - 4 = frac(k+1)(5k + 2)2.
endalign*
Let's see if we can generate this above expression, using what we know. We will, of course, use our induction hypothesis in the process.
beginalign*
sumlimits_i=1^k+1 5i - 4 & = sumlimits_i=1^k 5i - 4 + 5(k+1) - 4 & & textRules of summation \
& = frack(5k-3)2 + 5(k+1) - 4 & & textInduction Hypothesis \
& = frack(5k-3) + 10(k+1) - 82 \
& = frac5k^2 - 3k + 10k + 10 - 82 \
& = frac5k^2 + 7k + 22 \
& = fracleft(5k^2 + 5kright) + left(2k + 2right)2 \
& = frac5k(k+1) + 2(k+1)2 \
& = fracleft(5k + 2right)left(k+1right)2
endalign*
That's our goal. So, we proved the statement is true for $n = 1$, and that if it's true for $n = k$, it's true for $n = k + 1$. Therefore, the statement is true for all naturals $n geq 1$.

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edited Jul 20 at 1:03

answered Jul 20 at 0:57

Matt.P

671313

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up vote
0
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If $P(1)$ and $forall ngeq 1:[ P(n)to P(n+1)]$ then $forall ngeq 1 : P(n)$

Assume $sumlimits_j=1^k(5j-4)=dfrac(k(5k-3))2$, then we may derive:

$$beginalignsum_j=1^k+1 (5j-4)
&= left(sum_j=1^k(5j-4)right)+(5(k+1)-4)
\[2ex]&=frac(k(5k-3))2 + (5(k+1)-4)
\[2ex]& = frac(5k^2-3k)2+(5k+5-4)
\[2ex]&=frac(5k^2-3k+10k+10-8)2
\[2ex]&=frac(5k^2+10k+5-3k-3)2
\[2ex]&=frac(5(2k^2+2k+1)-3(k+1))2
\[2ex]&= frac(5(k+1)^2-3(k+1))2
\[2ex]&= frac(k+1)(5(k+1)-3)2
endalign$$

So we have that $sum_j=1^1(5j-4)= 1(5cdot 1-3)/2$ and for all $ngeq 1$, that $sum_k=1^n=n(5n-2)/2$ does imply that $sum_k=1^n+1=(n+1)(5(n+1)-3)/2$, and therefore: $$forall ngeq 1 ~:~sum_k=1^n=n(5n-2)/2$$

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edited Jul 20 at 1:32

answered Jul 20 at 0:46

Graham Kemp

80.1k43275

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote

I thought I would write out a answer that hopefully clarifies how you would approach this proof by induction.

The principle of induction states that, if we establish a 'base' case, usually $n = 0$ or $n = 1$ (depending on whether your class or textbook takes $0$ to be a natural or not, and depending on the particular statement you're trying to prove), and that it holds for some arbitrary natural number, $k$, then it holds for $k + 1$. Then, since you established $1$ as a basis, it holds for $2$, and since it holds for $2$, it holds for $3$, and so forth, until we cycle through all of the naturals. There are sometimes nuances in how to approach the base case, but this is the most general approach and all you need in this particular problem.

So, your goal is to prove that, for $n geq 1$, $sumlimits_i=1^n 5i - 4 = fracn(5n - 3)2$.

Let's start with the base case, which is $n = 1$ in this case. The left-hand side evaluates to $5(1) - 4 = 1$, while the right-hand side evaluates to $frac1(5(1) - 3)2 = 1$. So, this checks out.

Next, we assume that this statement is true for $n = k$, our 'induction hypothesis.' In other words, assume that
beginalign*
sumlimits_i=1^k 5i - 4 = frack(5k-3)2.
endalign*
We want to establish that the statement holds for $n = k + 1$. With a little bit of algebra, we can see that our goal is
beginalign*
sumlimits_i=1^k+1 5i - 4 = frac(k+1)(5k + 2)2.
endalign*
Let's see if we can generate this above expression, using what we know. We will, of course, use our induction hypothesis in the process.
beginalign*
sumlimits_i=1^k+1 5i - 4 & = sumlimits_i=1^k 5i - 4 + 5(k+1) - 4 & & textRules of summation \
& = frack(5k-3)2 + 5(k+1) - 4 & & textInduction Hypothesis \
& = frack(5k-3) + 10(k+1) - 82 \
& = frac5k^2 - 3k + 10k + 10 - 82 \
& = frac5k^2 + 7k + 22 \
& = fracleft(5k^2 + 5kright) + left(2k + 2right)2 \
& = frac5k(k+1) + 2(k+1)2 \
& = fracleft(5k + 2right)left(k+1right)2
endalign*
That's our goal. So, we proved the statement is true for $n = 1$, and that if it's true for $n = k$, it's true for $n = k + 1$. Therefore, the statement is true for all naturals $n geq 1$.

share|cite|improve this answer

edited Jul 20 at 1:03

answered Jul 20 at 0:57

Matt.P

671313

add a comment | 

up vote
1
down vote

I thought I would write out a answer that hopefully clarifies how you would approach this proof by induction.

The principle of induction states that, if we establish a 'base' case, usually $n = 0$ or $n = 1$ (depending on whether your class or textbook takes $0$ to be a natural or not, and depending on the particular statement you're trying to prove), and that it holds for some arbitrary natural number, $k$, then it holds for $k + 1$. Then, since you established $1$ as a basis, it holds for $2$, and since it holds for $2$, it holds for $3$, and so forth, until we cycle through all of the naturals. There are sometimes nuances in how to approach the base case, but this is the most general approach and all you need in this particular problem.

So, your goal is to prove that, for $n geq 1$, $sumlimits_i=1^n 5i - 4 = fracn(5n - 3)2$.

Let's start with the base case, which is $n = 1$ in this case. The left-hand side evaluates to $5(1) - 4 = 1$, while the right-hand side evaluates to $frac1(5(1) - 3)2 = 1$. So, this checks out.

Next, we assume that this statement is true for $n = k$, our 'induction hypothesis.' In other words, assume that
beginalign*
sumlimits_i=1^k 5i - 4 = frack(5k-3)2.
endalign*
We want to establish that the statement holds for $n = k + 1$. With a little bit of algebra, we can see that our goal is
beginalign*
sumlimits_i=1^k+1 5i - 4 = frac(k+1)(5k + 2)2.
endalign*
Let's see if we can generate this above expression, using what we know. We will, of course, use our induction hypothesis in the process.
beginalign*
sumlimits_i=1^k+1 5i - 4 & = sumlimits_i=1^k 5i - 4 + 5(k+1) - 4 & & textRules of summation \
& = frack(5k-3)2 + 5(k+1) - 4 & & textInduction Hypothesis \
& = frack(5k-3) + 10(k+1) - 82 \
& = frac5k^2 - 3k + 10k + 10 - 82 \
& = frac5k^2 + 7k + 22 \
& = fracleft(5k^2 + 5kright) + left(2k + 2right)2 \
& = frac5k(k+1) + 2(k+1)2 \
& = fracleft(5k + 2right)left(k+1right)2
endalign*
That's our goal. So, we proved the statement is true for $n = 1$, and that if it's true for $n = k$, it's true for $n = k + 1$. Therefore, the statement is true for all naturals $n geq 1$.

share|cite|improve this answer

edited Jul 20 at 1:03

answered Jul 20 at 0:57

Matt.P

671313

add a comment | 

up vote
1
down vote

up vote
1
down vote

I thought I would write out a answer that hopefully clarifies how you would approach this proof by induction.

The principle of induction states that, if we establish a 'base' case, usually $n = 0$ or $n = 1$ (depending on whether your class or textbook takes $0$ to be a natural or not, and depending on the particular statement you're trying to prove), and that it holds for some arbitrary natural number, $k$, then it holds for $k + 1$. Then, since you established $1$ as a basis, it holds for $2$, and since it holds for $2$, it holds for $3$, and so forth, until we cycle through all of the naturals. There are sometimes nuances in how to approach the base case, but this is the most general approach and all you need in this particular problem.

So, your goal is to prove that, for $n geq 1$, $sumlimits_i=1^n 5i - 4 = fracn(5n - 3)2$.

Let's start with the base case, which is $n = 1$ in this case. The left-hand side evaluates to $5(1) - 4 = 1$, while the right-hand side evaluates to $frac1(5(1) - 3)2 = 1$. So, this checks out.

Next, we assume that this statement is true for $n = k$, our 'induction hypothesis.' In other words, assume that
beginalign*
sumlimits_i=1^k 5i - 4 = frack(5k-3)2.
endalign*
We want to establish that the statement holds for $n = k + 1$. With a little bit of algebra, we can see that our goal is
beginalign*
sumlimits_i=1^k+1 5i - 4 = frac(k+1)(5k + 2)2.
endalign*
Let's see if we can generate this above expression, using what we know. We will, of course, use our induction hypothesis in the process.
beginalign*
sumlimits_i=1^k+1 5i - 4 & = sumlimits_i=1^k 5i - 4 + 5(k+1) - 4 & & textRules of summation \
& = frack(5k-3)2 + 5(k+1) - 4 & & textInduction Hypothesis \
& = frack(5k-3) + 10(k+1) - 82 \
& = frac5k^2 - 3k + 10k + 10 - 82 \
& = frac5k^2 + 7k + 22 \
& = fracleft(5k^2 + 5kright) + left(2k + 2right)2 \
& = frac5k(k+1) + 2(k+1)2 \
& = fracleft(5k + 2right)left(k+1right)2
endalign*
That's our goal. So, we proved the statement is true for $n = 1$, and that if it's true for $n = k$, it's true for $n = k + 1$. Therefore, the statement is true for all naturals $n geq 1$.

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edited Jul 20 at 1:03

answered Jul 20 at 0:57

Matt.P

671313

I thought I would write out a answer that hopefully clarifies how you would approach this proof by induction.

The principle of induction states that, if we establish a 'base' case, usually $n = 0$ or $n = 1$ (depending on whether your class or textbook takes $0$ to be a natural or not, and depending on the particular statement you're trying to prove), and that it holds for some arbitrary natural number, $k$, then it holds for $k + 1$. Then, since you established $1$ as a basis, it holds for $2$, and since it holds for $2$, it holds for $3$, and so forth, until we cycle through all of the naturals. There are sometimes nuances in how to approach the base case, but this is the most general approach and all you need in this particular problem.

So, your goal is to prove that, for $n geq 1$, $sumlimits_i=1^n 5i - 4 = fracn(5n - 3)2$.

Let's start with the base case, which is $n = 1$ in this case. The left-hand side evaluates to $5(1) - 4 = 1$, while the right-hand side evaluates to $frac1(5(1) - 3)2 = 1$. So, this checks out.

Next, we assume that this statement is true for $n = k$, our 'induction hypothesis.' In other words, assume that
beginalign*
sumlimits_i=1^k 5i - 4 = frack(5k-3)2.
endalign*
We want to establish that the statement holds for $n = k + 1$. With a little bit of algebra, we can see that our goal is
beginalign*
sumlimits_i=1^k+1 5i - 4 = frac(k+1)(5k + 2)2.
endalign*
Let's see if we can generate this above expression, using what we know. We will, of course, use our induction hypothesis in the process.
beginalign*
sumlimits_i=1^k+1 5i - 4 & = sumlimits_i=1^k 5i - 4 + 5(k+1) - 4 & & textRules of summation \
& = frack(5k-3)2 + 5(k+1) - 4 & & textInduction Hypothesis \
& = frack(5k-3) + 10(k+1) - 82 \
& = frac5k^2 - 3k + 10k + 10 - 82 \
& = frac5k^2 + 7k + 22 \
& = fracleft(5k^2 + 5kright) + left(2k + 2right)2 \
& = frac5k(k+1) + 2(k+1)2 \
& = fracleft(5k + 2right)left(k+1right)2
endalign*
That's our goal. So, we proved the statement is true for $n = 1$, and that if it's true for $n = k$, it's true for $n = k + 1$. Therefore, the statement is true for all naturals $n geq 1$.

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edited Jul 20 at 1:03

answered Jul 20 at 0:57

Matt.P

671313

share|cite|improve this answer

share|cite|improve this answer

edited Jul 20 at 1:03

edited Jul 20 at 1:03

edited Jul 20 at 1:03

answered Jul 20 at 0:57

Matt.P

671313

answered Jul 20 at 0:57

Matt.P

671313

answered Jul 20 at 0:57

Matt.P

671313

671313

add a comment | 

add a comment | 

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If $P(1)$ and $forall ngeq 1:[ P(n)to P(n+1)]$ then $forall ngeq 1 : P(n)$

Assume $sumlimits_j=1^k(5j-4)=dfrac(k(5k-3))2$, then we may derive:

$$beginalignsum_j=1^k+1 (5j-4)
&= left(sum_j=1^k(5j-4)right)+(5(k+1)-4)
\[2ex]&=frac(k(5k-3))2 + (5(k+1)-4)
\[2ex]& = frac(5k^2-3k)2+(5k+5-4)
\[2ex]&=frac(5k^2-3k+10k+10-8)2
\[2ex]&=frac(5k^2+10k+5-3k-3)2
\[2ex]&=frac(5(2k^2+2k+1)-3(k+1))2
\[2ex]&= frac(5(k+1)^2-3(k+1))2
\[2ex]&= frac(k+1)(5(k+1)-3)2
endalign$$

So we have that $sum_j=1^1(5j-4)= 1(5cdot 1-3)/2$ and for all $ngeq 1$, that $sum_k=1^n=n(5n-2)/2$ does imply that $sum_k=1^n+1=(n+1)(5(n+1)-3)/2$, and therefore: $$forall ngeq 1 ~:~sum_k=1^n=n(5n-2)/2$$

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edited Jul 20 at 1:32

answered Jul 20 at 0:46

Graham Kemp

80.1k43275

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up vote
0
down vote

If $P(1)$ and $forall ngeq 1:[ P(n)to P(n+1)]$ then $forall ngeq 1 : P(n)$

Assume $sumlimits_j=1^k(5j-4)=dfrac(k(5k-3))2$, then we may derive:

$$beginalignsum_j=1^k+1 (5j-4)
&= left(sum_j=1^k(5j-4)right)+(5(k+1)-4)
\[2ex]&=frac(k(5k-3))2 + (5(k+1)-4)
\[2ex]& = frac(5k^2-3k)2+(5k+5-4)
\[2ex]&=frac(5k^2-3k+10k+10-8)2
\[2ex]&=frac(5k^2+10k+5-3k-3)2
\[2ex]&=frac(5(2k^2+2k+1)-3(k+1))2
\[2ex]&= frac(5(k+1)^2-3(k+1))2
\[2ex]&= frac(k+1)(5(k+1)-3)2
endalign$$

So we have that $sum_j=1^1(5j-4)= 1(5cdot 1-3)/2$ and for all $ngeq 1$, that $sum_k=1^n=n(5n-2)/2$ does imply that $sum_k=1^n+1=(n+1)(5(n+1)-3)/2$, and therefore: $$forall ngeq 1 ~:~sum_k=1^n=n(5n-2)/2$$

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edited Jul 20 at 1:32

answered Jul 20 at 0:46

Graham Kemp

80.1k43275

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up vote
0
down vote

up vote
0
down vote

If $P(1)$ and $forall ngeq 1:[ P(n)to P(n+1)]$ then $forall ngeq 1 : P(n)$

Assume $sumlimits_j=1^k(5j-4)=dfrac(k(5k-3))2$, then we may derive:

$$beginalignsum_j=1^k+1 (5j-4)
&= left(sum_j=1^k(5j-4)right)+(5(k+1)-4)
\[2ex]&=frac(k(5k-3))2 + (5(k+1)-4)
\[2ex]& = frac(5k^2-3k)2+(5k+5-4)
\[2ex]&=frac(5k^2-3k+10k+10-8)2
\[2ex]&=frac(5k^2+10k+5-3k-3)2
\[2ex]&=frac(5(2k^2+2k+1)-3(k+1))2
\[2ex]&= frac(5(k+1)^2-3(k+1))2
\[2ex]&= frac(k+1)(5(k+1)-3)2
endalign$$

So we have that $sum_j=1^1(5j-4)= 1(5cdot 1-3)/2$ and for all $ngeq 1$, that $sum_k=1^n=n(5n-2)/2$ does imply that $sum_k=1^n+1=(n+1)(5(n+1)-3)/2$, and therefore: $$forall ngeq 1 ~:~sum_k=1^n=n(5n-2)/2$$

share|cite|improve this answer

edited Jul 20 at 1:32

answered Jul 20 at 0:46

Graham Kemp

80.1k43275

If $P(1)$ and $forall ngeq 1:[ P(n)to P(n+1)]$ then $forall ngeq 1 : P(n)$

Assume $sumlimits_j=1^k(5j-4)=dfrac(k(5k-3))2$, then we may derive:

$$beginalignsum_j=1^k+1 (5j-4)
&= left(sum_j=1^k(5j-4)right)+(5(k+1)-4)
\[2ex]&=frac(k(5k-3))2 + (5(k+1)-4)
\[2ex]& = frac(5k^2-3k)2+(5k+5-4)
\[2ex]&=frac(5k^2-3k+10k+10-8)2
\[2ex]&=frac(5k^2+10k+5-3k-3)2
\[2ex]&=frac(5(2k^2+2k+1)-3(k+1))2
\[2ex]&= frac(5(k+1)^2-3(k+1))2
\[2ex]&= frac(k+1)(5(k+1)-3)2
endalign$$

So we have that $sum_j=1^1(5j-4)= 1(5cdot 1-3)/2$ and for all $ngeq 1$, that $sum_k=1^n=n(5n-2)/2$ does imply that $sum_k=1^n+1=(n+1)(5(n+1)-3)/2$, and therefore: $$forall ngeq 1 ~:~sum_k=1^n=n(5n-2)/2$$

share|cite|improve this answer

edited Jul 20 at 1:32

answered Jul 20 at 0:46

Graham Kemp

80.1k43275

share|cite|improve this answer

share|cite|improve this answer

edited Jul 20 at 1:32

edited Jul 20 at 1:32

edited Jul 20 at 1:32

answered Jul 20 at 0:46

Graham Kemp

80.1k43275

answered Jul 20 at 0:46

Graham Kemp

80.1k43275

answered Jul 20 at 0:46

Graham Kemp

80.1k43275

80.1k43275

add a comment | 

add a comment |