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Having trouble with a Binomial proof by mathematical induction question: $sum_j=3^n binomj-12 = binomn3$ [closed]
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I can't work out how to prove this equation is true by proof of mathematical
Use mathematical induction to prove that, for $n ge 3$
$$sum_j=3^n binomj-12 = binomn3$$
Please help, thanks
summation induction binomial-coefficients binomial-theorem
edited Jul 20 at 8:05
Martin Sleziak
43.5k6113259
asked Jul 20 at 2:19
Harry Michael Kirwan
32
closed as off-topic by JMoravitz, Clement C., Leucippus, Shailesh, amWhy Jul 20 at 11:40
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Clement C., Leucippus, Shailesh, amWhy
add a comment |Â
up vote
-2
down vote
favorite
I can't work out how to prove this equation is true by proof of mathematical
Use mathematical induction to prove that, for $n ge 3$
$$sum_j=3^n binomj-12 = binomn3$$
Please help, thanks
summation induction binomial-coefficients binomial-theorem
edited Jul 20 at 8:05
Martin Sleziak
43.5k6113259
asked Jul 20 at 2:19
Harry Michael Kirwan
32
closed as off-topic by JMoravitz, Clement C., Leucippus, Shailesh, amWhy Jul 20 at 11:40
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Clement C., Leucippus, Shailesh, amWhy
An easier combinatorial argument: Count how many three-element subsets there are of $1,2,3,dots,n$ 1) Directly 2) By breaking into cases based on the largest element appearing.
– JMoravitz
Jul 20 at 2:23
Related: Hockey Stick Identity.
– JMoravitz
Jul 20 at 2:24
Possible duplicate of Proof of the Hockey-Stick Identity: $sumlimits_t=0^n binom tk = binomn+1k+1$
– JMoravitz
Jul 20 at 2:24
The original question did ask for you to prove something similar to Hockey-Stick identity. $binomnr + binomnr+1 = binomn+1r+1$ before this equation
– Harry Michael Kirwan
Jul 20 at 3:21
Some other related posts: Evaluate $sumlimits_k=1^n k^2$ and $sumlimits_k=1^n k(k+1)$ combinatorially and Simplify triangular sum of triangular numbers: $sum_i=1^n(frac12i(i+1))$. Also other posts linked to the latter might be of interest.
– Martin Sleziak
Jul 20 at 8:08
add a comment |Â
up vote
-2
down vote
favorite
up vote
-2
down vote
favorite
I can't work out how to prove this equation is true by proof of mathematical
Use mathematical induction to prove that, for $n ge 3$
$$sum_j=3^n binomj-12 = binomn3$$
Please help, thanks
summation induction binomial-coefficients binomial-theorem
edited Jul 20 at 8:05
Martin Sleziak
43.5k6113259
asked Jul 20 at 2:19
Harry Michael Kirwan
32
I can't work out how to prove this equation is true by proof of mathematical
Use mathematical induction to prove that, for $n ge 3$
$$sum_j=3^n binomj-12 = binomn3$$
Please help, thanks
summation induction binomial-coefficients binomial-theorem
edited Jul 20 at 8:05
Martin Sleziak
43.5k6113259
asked Jul 20 at 2:19
Harry Michael Kirwan
32
edited Jul 20 at 8:05
Martin Sleziak
43.5k6113259
edited Jul 20 at 8:05
Martin Sleziak
43.5k6113259
edited Jul 20 at 8:05
Martin Sleziak
43.5k6113259
43.5k6113259
asked Jul 20 at 2:19
Harry Michael Kirwan
32
asked Jul 20 at 2:19
Harry Michael Kirwan
32
asked Jul 20 at 2:19
Harry Michael Kirwan
32
32
closed as off-topic by JMoravitz, Clement C., Leucippus, Shailesh, amWhy Jul 20 at 11:40
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Clement C., Leucippus, Shailesh, amWhy
closed as off-topic by JMoravitz, Clement C., Leucippus, Shailesh, amWhy Jul 20 at 11:40
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Clement C., Leucippus, Shailesh, amWhy
An easier combinatorial argument: Count how many three-element subsets there are of $1,2,3,dots,n$ 1) Directly 2) By breaking into cases based on the largest element appearing.
– JMoravitz
Jul 20 at 2:23
Related: Hockey Stick Identity.
– JMoravitz
Jul 20 at 2:24
Possible duplicate of Proof of the Hockey-Stick Identity: $sumlimits_t=0^n binom tk = binomn+1k+1$
– JMoravitz
Jul 20 at 2:24
The original question did ask for you to prove something similar to Hockey-Stick identity. $binomnr + binomnr+1 = binomn+1r+1$ before this equation
– Harry Michael Kirwan
Jul 20 at 3:21
Some other related posts: Evaluate $sumlimits_k=1^n k^2$ and $sumlimits_k=1^n k(k+1)$ combinatorially and Simplify triangular sum of triangular numbers: $sum_i=1^n(frac12i(i+1))$. Also other posts linked to the latter might be of interest.
– Martin Sleziak
Jul 20 at 8:08
add a comment |Â
An easier combinatorial argument: Count how many three-element subsets there are of $1,2,3,dots,n$ 1) Directly 2) By breaking into cases based on the largest element appearing.
– JMoravitz
Jul 20 at 2:23
Related: Hockey Stick Identity.
– JMoravitz
Jul 20 at 2:24
Possible duplicate of Proof of the Hockey-Stick Identity: $sumlimits_t=0^n binom tk = binomn+1k+1$
– JMoravitz
Jul 20 at 2:24
The original question did ask for you to prove something similar to Hockey-Stick identity. $binomnr + binomnr+1 = binomn+1r+1$ before this equation
– Harry Michael Kirwan
Jul 20 at 3:21
Some other related posts: Evaluate $sumlimits_k=1^n k^2$ and $sumlimits_k=1^n k(k+1)$ combinatorially and Simplify triangular sum of triangular numbers: $sum_i=1^n(frac12i(i+1))$. Also other posts linked to the latter might be of interest.
– Martin Sleziak
Jul 20 at 8:08
An easier combinatorial argument: Count how many three-element subsets there are of $1,2,3,dots,n$ 1) Directly 2) By breaking into cases based on the largest element appearing.
– JMoravitz
Jul 20 at 2:23
An easier combinatorial argument: Count how many three-element subsets there are of $1,2,3,dots,n$ 1) Directly 2) By breaking into cases based on the largest element appearing.
– JMoravitz
Jul 20 at 2:23
Related: Hockey Stick Identity.
– JMoravitz
Jul 20 at 2:24
Related: Hockey Stick Identity.
– JMoravitz
Jul 20 at 2:24
Possible duplicate of Proof of the Hockey-Stick Identity: $sumlimits_t=0^n binom tk = binomn+1k+1$
– JMoravitz
Jul 20 at 2:24
Possible duplicate of Proof of the Hockey-Stick Identity: $sumlimits_t=0^n binom tk = binomn+1k+1$
– JMoravitz
Jul 20 at 2:24
The original question did ask for you to prove something similar to Hockey-Stick identity. $binomnr + binomnr+1 = binomn+1r+1$ before this equation
– Harry Michael Kirwan
Jul 20 at 3:21
The original question did ask for you to prove something similar to Hockey-Stick identity. $binomnr + binomnr+1 = binomn+1r+1$ before this equation
– Harry Michael Kirwan
Jul 20 at 3:21
Some other related posts: Evaluate $sumlimits_k=1^n k^2$ and $sumlimits_k=1^n k(k+1)$ combinatorially and Simplify triangular sum of triangular numbers: $sum_i=1^n(frac12i(i+1))$. Also other posts linked to the latter might be of interest.
– Martin Sleziak
Jul 20 at 8:08
Some other related posts: Evaluate $sumlimits_k=1^n k^2$ and $sumlimits_k=1^n k(k+1)$ combinatorially and Simplify triangular sum of triangular numbers: $sum_i=1^n(frac12i(i+1))$. Also other posts linked to the latter might be of interest.
– Martin Sleziak
Jul 20 at 8:08
add a comment |Â
1 Answer
1
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oldest
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up vote
0
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Assume $$sum_j=3^n binomj-12 = binomn3$$ is true. Let's prove
$$sum_j=3^n+1 binomj-12 = binomn+13$$
So
$$sum_j=3^n+1 binomj-12 = sum_j=3^n binomj-12 + binomn2 = binomn3+ binomn2 = fracn!3!(n-3)!+fracn!2!(n-2)! = frac(n+1)!3!(n-2)! = binomn+13$$
answered Jul 20 at 2:38
Ahmad Bazzi
2,6271417
You can't assume true for the original equation, I'm trying to prove that the original equation is true.
– Harry Michael Kirwan
Jul 20 at 3:18
1
@HarryMichaelKirwan that's how induction works. You assume that it is true for $n$ and prove that it being true for $n$, implies it is true for $n+1$. You must also prove that it is true for smallest possible value of $n$. In this case that is $n=3$.
– Piyush Divyanakar
Jul 20 at 3:28
@HarryMichaelKirwan, I could have told you it is true for $n-1$, then i'd do the same steps to prove it for $n$. Try it yourself !!!
– Ahmad Bazzi
Jul 20 at 3:31
@AhmadBazzi actually yeah my bad sorry, but what I don't understand is where you got the $binomn2$ going from $sum_j=3^n+1 binomj-12 $
– Harry Michael Kirwan
Jul 20 at 3:33
what is $binomj-12$ computed at $j = n+1$ ?
– Ahmad Bazzi
Jul 20 at 3:38
 |Â
show 1 more comment
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
Assume $$sum_j=3^n binomj-12 = binomn3$$ is true. Let's prove
$$sum_j=3^n+1 binomj-12 = binomn+13$$
So
$$sum_j=3^n+1 binomj-12 = sum_j=3^n binomj-12 + binomn2 = binomn3+ binomn2 = fracn!3!(n-3)!+fracn!2!(n-2)! = frac(n+1)!3!(n-2)! = binomn+13$$
answered Jul 20 at 2:38
Ahmad Bazzi
2,6271417
You can't assume true for the original equation, I'm trying to prove that the original equation is true.
– Harry Michael Kirwan
Jul 20 at 3:18
1
@HarryMichaelKirwan that's how induction works. You assume that it is true for $n$ and prove that it being true for $n$, implies it is true for $n+1$. You must also prove that it is true for smallest possible value of $n$. In this case that is $n=3$.
– Piyush Divyanakar
Jul 20 at 3:28
@HarryMichaelKirwan, I could have told you it is true for $n-1$, then i'd do the same steps to prove it for $n$. Try it yourself !!!
– Ahmad Bazzi
Jul 20 at 3:31
@AhmadBazzi actually yeah my bad sorry, but what I don't understand is where you got the $binomn2$ going from $sum_j=3^n+1 binomj-12 $
– Harry Michael Kirwan
Jul 20 at 3:33
what is $binomj-12$ computed at $j = n+1$ ?
– Ahmad Bazzi
Jul 20 at 3:38
 |Â
show 1 more comment
up vote
0
down vote
accepted
Assume $$sum_j=3^n binomj-12 = binomn3$$ is true. Let's prove
$$sum_j=3^n+1 binomj-12 = binomn+13$$
So
$$sum_j=3^n+1 binomj-12 = sum_j=3^n binomj-12 + binomn2 = binomn3+ binomn2 = fracn!3!(n-3)!+fracn!2!(n-2)! = frac(n+1)!3!(n-2)! = binomn+13$$
answered Jul 20 at 2:38
Ahmad Bazzi
2,6271417
You can't assume true for the original equation, I'm trying to prove that the original equation is true.
– Harry Michael Kirwan
Jul 20 at 3:18
1
@HarryMichaelKirwan that's how induction works. You assume that it is true for $n$ and prove that it being true for $n$, implies it is true for $n+1$. You must also prove that it is true for smallest possible value of $n$. In this case that is $n=3$.
– Piyush Divyanakar
Jul 20 at 3:28
@HarryMichaelKirwan, I could have told you it is true for $n-1$, then i'd do the same steps to prove it for $n$. Try it yourself !!!
– Ahmad Bazzi
Jul 20 at 3:31
@AhmadBazzi actually yeah my bad sorry, but what I don't understand is where you got the $binomn2$ going from $sum_j=3^n+1 binomj-12 $
– Harry Michael Kirwan
Jul 20 at 3:33
what is $binomj-12$ computed at $j = n+1$ ?
– Ahmad Bazzi
Jul 20 at 3:38
 |Â
show 1 more comment
up vote
0
down vote
accepted
up vote
0
down vote
accepted
Assume $$sum_j=3^n binomj-12 = binomn3$$ is true. Let's prove
$$sum_j=3^n+1 binomj-12 = binomn+13$$
So
$$sum_j=3^n+1 binomj-12 = sum_j=3^n binomj-12 + binomn2 = binomn3+ binomn2 = fracn!3!(n-3)!+fracn!2!(n-2)! = frac(n+1)!3!(n-2)! = binomn+13$$
answered Jul 20 at 2:38
Ahmad Bazzi
2,6271417
Assume $$sum_j=3^n binomj-12 = binomn3$$ is true. Let's prove
$$sum_j=3^n+1 binomj-12 = binomn+13$$
So
$$sum_j=3^n+1 binomj-12 = sum_j=3^n binomj-12 + binomn2 = binomn3+ binomn2 = fracn!3!(n-3)!+fracn!2!(n-2)! = frac(n+1)!3!(n-2)! = binomn+13$$
answered Jul 20 at 2:38
Ahmad Bazzi
2,6271417
answered Jul 20 at 2:38
Ahmad Bazzi
2,6271417
answered Jul 20 at 2:38
Ahmad Bazzi
2,6271417
answered Jul 20 at 2:38
Ahmad Bazzi
2,6271417
2,6271417
You can't assume true for the original equation, I'm trying to prove that the original equation is true.
– Harry Michael Kirwan
Jul 20 at 3:18
1
@HarryMichaelKirwan that's how induction works. You assume that it is true for $n$ and prove that it being true for $n$, implies it is true for $n+1$. You must also prove that it is true for smallest possible value of $n$. In this case that is $n=3$.
– Piyush Divyanakar
Jul 20 at 3:28
@HarryMichaelKirwan, I could have told you it is true for $n-1$, then i'd do the same steps to prove it for $n$. Try it yourself !!!
– Ahmad Bazzi
Jul 20 at 3:31
@AhmadBazzi actually yeah my bad sorry, but what I don't understand is where you got the $binomn2$ going from $sum_j=3^n+1 binomj-12 $
– Harry Michael Kirwan
Jul 20 at 3:33
what is $binomj-12$ computed at $j = n+1$ ?
– Ahmad Bazzi
Jul 20 at 3:38
 |Â
show 1 more comment
You can't assume true for the original equation, I'm trying to prove that the original equation is true.
– Harry Michael Kirwan
Jul 20 at 3:18
1
@HarryMichaelKirwan that's how induction works. You assume that it is true for $n$ and prove that it being true for $n$, implies it is true for $n+1$. You must also prove that it is true for smallest possible value of $n$. In this case that is $n=3$.
– Piyush Divyanakar
Jul 20 at 3:28
@HarryMichaelKirwan, I could have told you it is true for $n-1$, then i'd do the same steps to prove it for $n$. Try it yourself !!!
– Ahmad Bazzi
Jul 20 at 3:31
@AhmadBazzi actually yeah my bad sorry, but what I don't understand is where you got the $binomn2$ going from $sum_j=3^n+1 binomj-12 $
– Harry Michael Kirwan
Jul 20 at 3:33
what is $binomj-12$ computed at $j = n+1$ ?
– Ahmad Bazzi
Jul 20 at 3:38
You can't assume true for the original equation, I'm trying to prove that the original equation is true.
– Harry Michael Kirwan
Jul 20 at 3:18
You can't assume true for the original equation, I'm trying to prove that the original equation is true.
– Harry Michael Kirwan
Jul 20 at 3:18
1
1
@HarryMichaelKirwan that's how induction works. You assume that it is true for $n$ and prove that it being true for $n$, implies it is true for $n+1$. You must also prove that it is true for smallest possible value of $n$. In this case that is $n=3$.
– Piyush Divyanakar
Jul 20 at 3:28
@HarryMichaelKirwan that's how induction works. You assume that it is true for $n$ and prove that it being true for $n$, implies it is true for $n+1$. You must also prove that it is true for smallest possible value of $n$. In this case that is $n=3$.
– Piyush Divyanakar
Jul 20 at 3:28
@HarryMichaelKirwan, I could have told you it is true for $n-1$, then i'd do the same steps to prove it for $n$. Try it yourself !!!
– Ahmad Bazzi
Jul 20 at 3:31
@HarryMichaelKirwan, I could have told you it is true for $n-1$, then i'd do the same steps to prove it for $n$. Try it yourself !!!
– Ahmad Bazzi
Jul 20 at 3:31
@AhmadBazzi actually yeah my bad sorry, but what I don't understand is where you got the $binomn2$ going from $sum_j=3^n+1 binomj-12 $
– Harry Michael Kirwan
Jul 20 at 3:33
@AhmadBazzi actually yeah my bad sorry, but what I don't understand is where you got the $binomn2$ going from $sum_j=3^n+1 binomj-12 $
– Harry Michael Kirwan
Jul 20 at 3:33
what is $binomj-12$ computed at $j = n+1$ ?
– Ahmad Bazzi
Jul 20 at 3:38
what is $binomj-12$ computed at $j = n+1$ ?
– Ahmad Bazzi
Jul 20 at 3:38
 |Â
show 1 more comment
An easier combinatorial argument: Count how many three-element subsets there are of $1,2,3,dots,n$ 1) Directly 2) By breaking into cases based on the largest element appearing.
– JMoravitz
Jul 20 at 2:23
Related: Hockey Stick Identity.
– JMoravitz
Jul 20 at 2:24
Possible duplicate of Proof of the Hockey-Stick Identity: $sumlimits_t=0^n binom tk = binomn+1k+1$
– JMoravitz
Jul 20 at 2:24
The original question did ask for you to prove something similar to Hockey-Stick identity. $binomnr + binomnr+1 = binomn+1r+1$ before this equation
– Harry Michael Kirwan
Jul 20 at 3:21
Some other related posts: Evaluate $sumlimits_k=1^n k^2$ and $sumlimits_k=1^n k(k+1)$ combinatorially and Simplify triangular sum of triangular numbers: $sum_i=1^n(frac12i(i+1))$. Also other posts linked to the latter might be of interest.
– Martin Sleziak
Jul 20 at 8:08