Recent content by jenny Downer
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Proving the Binomial Theorem: Simplifying 3^k C(n,k) = 2^2n
Use the Binomial Theorem to show that n summation k-0 3^k C(n,k) = 2^2n hint of the question is 3^k C(n,k) = 3^k 1^n-k C(n,k)- jenny Downer
- Thread
- Binomial Binomial theorem Theorem
- Replies: 1
- Forum: Precalculus Mathematics Homework Help
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J
Using the Binomial Theorem to Show 3^k C(n,k) = 3^k 1^n-k C(n,k)
It should be n sum 3^k C(n,k) = 2^2n k=0 the hint is 3^k C(n,k) = 3^k 1^n-k C(n,k)- jenny Downer
- Post #4
- Forum: Precalculus Mathematics Homework Help
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J
Using the Binomial Theorem to Show 3^k C(n,k) = 3^k 1^n-k C(n,k)
i wanted to explain a bit more- jenny Downer
- Post #3
- Forum: Precalculus Mathematics Homework Help
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J
Using the Binomial Theorem to Show 3^k C(n,k) = 3^k 1^n-k C(n,k)
how do i use binomial to show that 3^k C(n,k) = 3^k 1^n-k C(n,k)- jenny Downer
- Thread
- Binomial Binomial theorem Theorem
- Replies: 4
- Forum: Precalculus Mathematics Homework Help