Mixing
My induction does not prove my conjecture. Is my conjecture wrong or is my induction insufficient/wrong?
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I am asked to consider the following sequence:
$$1=0+1\2+3+4=1+8\5+6+7+8+9=8+27\10+11+12+13+14+15+16=27+64$$
The questions are:
a) What is the next equality in this sequence?
b) What conjecture is suggested by these equalities?
c) Prove the conjecture in (b) by induction.
Here are my answers:
a) $17+18+19+20+21+22+23+24+25=27+64$
b) $forall i in mathbbZ^+, sum_k=(i-1)^2+1^i^2k=(i-1)(i-1)^2+(i)^3$
c) $$textProof: We proceed by induction. \textBase case: sum_k=(1-1)^2+1^1^2k=1, (1-1)(1-1)^2+(1)^3=0+1=1, \textThus: LHS=RHS \textWe assume for some integer z, sum_k=(z-1)^2+1^z^2k=(z-1)(z-1)^2+(z)^3 \textWe need to show, sum_k=(z-1)^2+1^(z+1)^2k=(z)^3+(z+1)^3 \textHence, sum_k=(z-1)^2+1^z^2k + (z+1)^2 = sum_k=(z-1)^2+1^(z+1)^2k=(z-1)(z-1)^2+(z)^3+(z+1)^2$$
But from there I can't get:
$$(z-1)(z-1)^2+(z)^3+(z+1)^2=(z)^3+(z+1)^3$$
Maybe I am misunderstanding something here that should be obvious but I am still learning to prove by induction. Using strong induction has gone through my mind but at this point I don't feel confident enough in my understanding of strong induction to use it here. (I am self-learning this.)
Update: This is how I proved it by using the index of the summation.
We notice in what we want to show, that if we change the indexing to zero, then we can write the summation as follows:
$$ textLet pge 0 text. \ sum_k=(p)^2+1^(p+1)^2k=((p+1)-1)((p+1)-1)^2+(p+1)^3 = (p)^3+(p+1)^3 blacksquare$$
proof-verification summation induction
asked Jul 22 at 13:18
Cro-Magnon
371112
add a comment |Â
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0
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I am asked to consider the following sequence:
$$1=0+1\2+3+4=1+8\5+6+7+8+9=8+27\10+11+12+13+14+15+16=27+64$$
The questions are:
a) What is the next equality in this sequence?
b) What conjecture is suggested by these equalities?
c) Prove the conjecture in (b) by induction.
Here are my answers:
a) $17+18+19+20+21+22+23+24+25=27+64$
b) $forall i in mathbbZ^+, sum_k=(i-1)^2+1^i^2k=(i-1)(i-1)^2+(i)^3$
c) $$textProof: We proceed by induction. \textBase case: sum_k=(1-1)^2+1^1^2k=1, (1-1)(1-1)^2+(1)^3=0+1=1, \textThus: LHS=RHS \textWe assume for some integer z, sum_k=(z-1)^2+1^z^2k=(z-1)(z-1)^2+(z)^3 \textWe need to show, sum_k=(z-1)^2+1^(z+1)^2k=(z)^3+(z+1)^3 \textHence, sum_k=(z-1)^2+1^z^2k + (z+1)^2 = sum_k=(z-1)^2+1^(z+1)^2k=(z-1)(z-1)^2+(z)^3+(z+1)^2$$
But from there I can't get:
$$(z-1)(z-1)^2+(z)^3+(z+1)^2=(z)^3+(z+1)^3$$
Maybe I am misunderstanding something here that should be obvious but I am still learning to prove by induction. Using strong induction has gone through my mind but at this point I don't feel confident enough in my understanding of strong induction to use it here. (I am self-learning this.)
Update: This is how I proved it by using the index of the summation.
We notice in what we want to show, that if we change the indexing to zero, then we can write the summation as follows:
$$ textLet pge 0 text. \ sum_k=(p)^2+1^(p+1)^2k=((p+1)-1)((p+1)-1)^2+(p+1)^3 = (p)^3+(p+1)^3 blacksquare$$
proof-verification summation induction
asked Jul 22 at 13:18
Cro-Magnon
371112
3
You've assumed it true for $$sum_k=(z-1)^2+1^z^2k$$. You should then follow with showing $$sum_k=z^2+1^(z+1)^2k$$ which I believe is where your error lies
– Rhys Hughes
Jul 22 at 13:22
1
If you have two sequences, $a_n,,b_n$ and you wish to show they are equal it suffices to show that $a_0=b_0$ and that $a_n-a_n-1=b_n-b_n-1$. However, in this case I think a direct method is much easier (see my posted solution).
– lulu
Jul 22 at 13:39
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am asked to consider the following sequence:
$$1=0+1\2+3+4=1+8\5+6+7+8+9=8+27\10+11+12+13+14+15+16=27+64$$
The questions are:
a) What is the next equality in this sequence?
b) What conjecture is suggested by these equalities?
c) Prove the conjecture in (b) by induction.
Here are my answers:
a) $17+18+19+20+21+22+23+24+25=27+64$
b) $forall i in mathbbZ^+, sum_k=(i-1)^2+1^i^2k=(i-1)(i-1)^2+(i)^3$
c) $$textProof: We proceed by induction. \textBase case: sum_k=(1-1)^2+1^1^2k=1, (1-1)(1-1)^2+(1)^3=0+1=1, \textThus: LHS=RHS \textWe assume for some integer z, sum_k=(z-1)^2+1^z^2k=(z-1)(z-1)^2+(z)^3 \textWe need to show, sum_k=(z-1)^2+1^(z+1)^2k=(z)^3+(z+1)^3 \textHence, sum_k=(z-1)^2+1^z^2k + (z+1)^2 = sum_k=(z-1)^2+1^(z+1)^2k=(z-1)(z-1)^2+(z)^3+(z+1)^2$$
But from there I can't get:
$$(z-1)(z-1)^2+(z)^3+(z+1)^2=(z)^3+(z+1)^3$$
Maybe I am misunderstanding something here that should be obvious but I am still learning to prove by induction. Using strong induction has gone through my mind but at this point I don't feel confident enough in my understanding of strong induction to use it here. (I am self-learning this.)
Update: This is how I proved it by using the index of the summation.
We notice in what we want to show, that if we change the indexing to zero, then we can write the summation as follows:
$$ textLet pge 0 text. \ sum_k=(p)^2+1^(p+1)^2k=((p+1)-1)((p+1)-1)^2+(p+1)^3 = (p)^3+(p+1)^3 blacksquare$$
proof-verification summation induction
asked Jul 22 at 13:18
Cro-Magnon
371112
I am asked to consider the following sequence:
$$1=0+1\2+3+4=1+8\5+6+7+8+9=8+27\10+11+12+13+14+15+16=27+64$$
The questions are:
a) What is the next equality in this sequence?
b) What conjecture is suggested by these equalities?
c) Prove the conjecture in (b) by induction.
Here are my answers:
a) $17+18+19+20+21+22+23+24+25=27+64$
b) $forall i in mathbbZ^+, sum_k=(i-1)^2+1^i^2k=(i-1)(i-1)^2+(i)^3$
c) $$textProof: We proceed by induction. \textBase case: sum_k=(1-1)^2+1^1^2k=1, (1-1)(1-1)^2+(1)^3=0+1=1, \textThus: LHS=RHS \textWe assume for some integer z, sum_k=(z-1)^2+1^z^2k=(z-1)(z-1)^2+(z)^3 \textWe need to show, sum_k=(z-1)^2+1^(z+1)^2k=(z)^3+(z+1)^3 \textHence, sum_k=(z-1)^2+1^z^2k + (z+1)^2 = sum_k=(z-1)^2+1^(z+1)^2k=(z-1)(z-1)^2+(z)^3+(z+1)^2$$
But from there I can't get:
$$(z-1)(z-1)^2+(z)^3+(z+1)^2=(z)^3+(z+1)^3$$
Maybe I am misunderstanding something here that should be obvious but I am still learning to prove by induction. Using strong induction has gone through my mind but at this point I don't feel confident enough in my understanding of strong induction to use it here. (I am self-learning this.)
Update: This is how I proved it by using the index of the summation.
We notice in what we want to show, that if we change the indexing to zero, then we can write the summation as follows:
$$ textLet pge 0 text. \ sum_k=(p)^2+1^(p+1)^2k=((p+1)-1)((p+1)-1)^2+(p+1)^3 = (p)^3+(p+1)^3 blacksquare$$
proof-verification summation induction
asked Jul 22 at 13:18
Cro-Magnon
371112
edited Jul 22 at 16:18
asked Jul 22 at 13:18
Cro-Magnon
371112
asked Jul 22 at 13:18
Cro-Magnon
371112
asked Jul 22 at 13:18
Cro-Magnon
371112
371112
3
You've assumed it true for $$sum_k=(z-1)^2+1^z^2k$$. You should then follow with showing $$sum_k=z^2+1^(z+1)^2k$$ which I believe is where your error lies
– Rhys Hughes
Jul 22 at 13:22
1
If you have two sequences, $a_n,,b_n$ and you wish to show they are equal it suffices to show that $a_0=b_0$ and that $a_n-a_n-1=b_n-b_n-1$. However, in this case I think a direct method is much easier (see my posted solution).
– lulu
Jul 22 at 13:39
add a comment |Â
3
You've assumed it true for $$sum_k=(z-1)^2+1^z^2k$$. You should then follow with showing $$sum_k=z^2+1^(z+1)^2k$$ which I believe is where your error lies
– Rhys Hughes
Jul 22 at 13:22
1
If you have two sequences, $a_n,,b_n$ and you wish to show they are equal it suffices to show that $a_0=b_0$ and that $a_n-a_n-1=b_n-b_n-1$. However, in this case I think a direct method is much easier (see my posted solution).
– lulu
Jul 22 at 13:39
3
3
You've assumed it true for $$sum_k=(z-1)^2+1^z^2k$$. You should then follow with showing $$sum_k=z^2+1^(z+1)^2k$$ which I believe is where your error lies
– Rhys Hughes
Jul 22 at 13:22
You've assumed it true for $$sum_k=(z-1)^2+1^z^2k$$. You should then follow with showing $$sum_k=z^2+1^(z+1)^2k$$ which I believe is where your error lies
– Rhys Hughes
Jul 22 at 13:22
1
1
If you have two sequences, $a_n,,b_n$ and you wish to show they are equal it suffices to show that $a_0=b_0$ and that $a_n-a_n-1=b_n-b_n-1$. However, in this case I think a direct method is much easier (see my posted solution).
– lulu
Jul 22 at 13:39
If you have two sequences, $a_n,,b_n$ and you wish to show they are equal it suffices to show that $a_0=b_0$ and that $a_n-a_n-1=b_n-b_n-1$. However, in this case I think a direct method is much easier (see my posted solution).
– lulu
Jul 22 at 13:39
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
3
down vote
accepted
Your approach is correct but you have some minor mistakes in your approach.
For example $$17+18+19+20+21+22+23+24+25=27+64$$ should have been
$$17+18+19+20+21+22+23+24+25=64+125$$
and $$textWe need to show, sum_k=(z-1)^2+1^(z+1)^2k=(z)^3+(z+1)^3$$
Should have been $$textWe need to show, sum_k=z^2+1^(z+1)^2k=(z)^3+(z+1)^3$$
answered Jul 22 at 13:43
Mohammad Riazi-Kermani
27.5k41852
I added an update with the final step of my proof. Could you take a look and see whether it is sufficient? Thanks so much!
– Cro-Magnon
Jul 22 at 16:20
add a comment |Â
up vote
2
down vote
In this case, I think a direct solution is easier than induction (though surely induction should work)
Index the formulas starting with $n=0$. The $n^th$ left hand sum can be written as: $$m(n)-n,m(n)-n+1,cdots,m(n)-1, m(n),m(n)+1,cdots, m(n)+n$$
for $m(n)=n^2+n+1$.
It follows that the sum of the terms in the $n^th$ sum is $$(2n+1)m(n)=(2n+1)(n^2+n+1)=2n^3+3n^2+3n+1$$
On the other hand the $n^th$ right hand is $$(n+1)^3+(n)^3=2n^3+3n^2+3n+1$$ and we are done.
answered Jul 22 at 13:33
lulu
35.1k14072
Could you take a look at my update? Your input is highly appreciated.
– Cro-Magnon
Jul 22 at 16:22
1
@Cro-Magnon The update looks good!
– lulu
Jul 22 at 16:39
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Your approach is correct but you have some minor mistakes in your approach.
For example $$17+18+19+20+21+22+23+24+25=27+64$$ should have been
$$17+18+19+20+21+22+23+24+25=64+125$$
and $$textWe need to show, sum_k=(z-1)^2+1^(z+1)^2k=(z)^3+(z+1)^3$$
Should have been $$textWe need to show, sum_k=z^2+1^(z+1)^2k=(z)^3+(z+1)^3$$
answered Jul 22 at 13:43
Mohammad Riazi-Kermani
27.5k41852
I added an update with the final step of my proof. Could you take a look and see whether it is sufficient? Thanks so much!
– Cro-Magnon
Jul 22 at 16:20
add a comment |Â
up vote
3
down vote
accepted
Your approach is correct but you have some minor mistakes in your approach.
For example $$17+18+19+20+21+22+23+24+25=27+64$$ should have been
$$17+18+19+20+21+22+23+24+25=64+125$$
and $$textWe need to show, sum_k=(z-1)^2+1^(z+1)^2k=(z)^3+(z+1)^3$$
Should have been $$textWe need to show, sum_k=z^2+1^(z+1)^2k=(z)^3+(z+1)^3$$
answered Jul 22 at 13:43
Mohammad Riazi-Kermani
27.5k41852
I added an update with the final step of my proof. Could you take a look and see whether it is sufficient? Thanks so much!
– Cro-Magnon
Jul 22 at 16:20
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Your approach is correct but you have some minor mistakes in your approach.
For example $$17+18+19+20+21+22+23+24+25=27+64$$ should have been
$$17+18+19+20+21+22+23+24+25=64+125$$
and $$textWe need to show, sum_k=(z-1)^2+1^(z+1)^2k=(z)^3+(z+1)^3$$
Should have been $$textWe need to show, sum_k=z^2+1^(z+1)^2k=(z)^3+(z+1)^3$$
answered Jul 22 at 13:43
Mohammad Riazi-Kermani
27.5k41852
Your approach is correct but you have some minor mistakes in your approach.
For example $$17+18+19+20+21+22+23+24+25=27+64$$ should have been
$$17+18+19+20+21+22+23+24+25=64+125$$
and $$textWe need to show, sum_k=(z-1)^2+1^(z+1)^2k=(z)^3+(z+1)^3$$
Should have been $$textWe need to show, sum_k=z^2+1^(z+1)^2k=(z)^3+(z+1)^3$$
answered Jul 22 at 13:43
Mohammad Riazi-Kermani
27.5k41852
answered Jul 22 at 13:43
Mohammad Riazi-Kermani
27.5k41852
answered Jul 22 at 13:43
Mohammad Riazi-Kermani
27.5k41852
answered Jul 22 at 13:43
Mohammad Riazi-Kermani
27.5k41852
27.5k41852
I added an update with the final step of my proof. Could you take a look and see whether it is sufficient? Thanks so much!
– Cro-Magnon
Jul 22 at 16:20
add a comment |Â
I added an update with the final step of my proof. Could you take a look and see whether it is sufficient? Thanks so much!
– Cro-Magnon
Jul 22 at 16:20
I added an update with the final step of my proof. Could you take a look and see whether it is sufficient? Thanks so much!
– Cro-Magnon
Jul 22 at 16:20
I added an update with the final step of my proof. Could you take a look and see whether it is sufficient? Thanks so much!
– Cro-Magnon
Jul 22 at 16:20
add a comment |Â
up vote
2
down vote
In this case, I think a direct solution is easier than induction (though surely induction should work)
Index the formulas starting with $n=0$. The $n^th$ left hand sum can be written as: $$m(n)-n,m(n)-n+1,cdots,m(n)-1, m(n),m(n)+1,cdots, m(n)+n$$
for $m(n)=n^2+n+1$.
It follows that the sum of the terms in the $n^th$ sum is $$(2n+1)m(n)=(2n+1)(n^2+n+1)=2n^3+3n^2+3n+1$$
On the other hand the $n^th$ right hand is $$(n+1)^3+(n)^3=2n^3+3n^2+3n+1$$ and we are done.
answered Jul 22 at 13:33
lulu
35.1k14072
Could you take a look at my update? Your input is highly appreciated.
– Cro-Magnon
Jul 22 at 16:22
1
@Cro-Magnon The update looks good!
– lulu
Jul 22 at 16:39
add a comment |Â
up vote
2
down vote
In this case, I think a direct solution is easier than induction (though surely induction should work)
Index the formulas starting with $n=0$. The $n^th$ left hand sum can be written as: $$m(n)-n,m(n)-n+1,cdots,m(n)-1, m(n),m(n)+1,cdots, m(n)+n$$
for $m(n)=n^2+n+1$.
It follows that the sum of the terms in the $n^th$ sum is $$(2n+1)m(n)=(2n+1)(n^2+n+1)=2n^3+3n^2+3n+1$$
On the other hand the $n^th$ right hand is $$(n+1)^3+(n)^3=2n^3+3n^2+3n+1$$ and we are done.
answered Jul 22 at 13:33
lulu
35.1k14072
Could you take a look at my update? Your input is highly appreciated.
– Cro-Magnon
Jul 22 at 16:22
1
@Cro-Magnon The update looks good!
– lulu
Jul 22 at 16:39
add a comment |Â
up vote
2
down vote
up vote
2
down vote
In this case, I think a direct solution is easier than induction (though surely induction should work)
Index the formulas starting with $n=0$. The $n^th$ left hand sum can be written as: $$m(n)-n,m(n)-n+1,cdots,m(n)-1, m(n),m(n)+1,cdots, m(n)+n$$
for $m(n)=n^2+n+1$.
It follows that the sum of the terms in the $n^th$ sum is $$(2n+1)m(n)=(2n+1)(n^2+n+1)=2n^3+3n^2+3n+1$$
On the other hand the $n^th$ right hand is $$(n+1)^3+(n)^3=2n^3+3n^2+3n+1$$ and we are done.
answered Jul 22 at 13:33
lulu
35.1k14072
In this case, I think a direct solution is easier than induction (though surely induction should work)
Index the formulas starting with $n=0$. The $n^th$ left hand sum can be written as: $$m(n)-n,m(n)-n+1,cdots,m(n)-1, m(n),m(n)+1,cdots, m(n)+n$$
for $m(n)=n^2+n+1$.
It follows that the sum of the terms in the $n^th$ sum is $$(2n+1)m(n)=(2n+1)(n^2+n+1)=2n^3+3n^2+3n+1$$
On the other hand the $n^th$ right hand is $$(n+1)^3+(n)^3=2n^3+3n^2+3n+1$$ and we are done.
answered Jul 22 at 13:33
lulu
35.1k14072
answered Jul 22 at 13:33
lulu
35.1k14072
answered Jul 22 at 13:33
lulu
35.1k14072
answered Jul 22 at 13:33
lulu
35.1k14072
35.1k14072
Could you take a look at my update? Your input is highly appreciated.
– Cro-Magnon
Jul 22 at 16:22
1
@Cro-Magnon The update looks good!
– lulu
Jul 22 at 16:39
add a comment |Â
Could you take a look at my update? Your input is highly appreciated.
– Cro-Magnon
Jul 22 at 16:22
1
@Cro-Magnon The update looks good!
– lulu
Jul 22 at 16:39
Could you take a look at my update? Your input is highly appreciated.
– Cro-Magnon
Jul 22 at 16:22
Could you take a look at my update? Your input is highly appreciated.
– Cro-Magnon
Jul 22 at 16:22
1
1
@Cro-Magnon The update looks good!
– lulu
Jul 22 at 16:39
@Cro-Magnon The update looks good!
– lulu
Jul 22 at 16:39
add a comment |Â
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3
You've assumed it true for $$sum_k=(z-1)^2+1^z^2k$$. You should then follow with showing $$sum_k=z^2+1^(z+1)^2k$$ which I believe is where your error lies
– Rhys Hughes
Jul 22 at 13:22
1
If you have two sequences, $a_n,,b_n$ and you wish to show they are equal it suffices to show that $a_0=b_0$ and that $a_n-a_n-1=b_n-b_n-1$. However, in this case I think a direct method is much easier (see my posted solution).
– lulu
Jul 22 at 13:39