Wednesday, April 27, 2016

algebra precalculus - Prove by induction sumni=1i3=fracn2(n+1)24 for nge1

Prove the following statement S(n) for n\ge1:



\sum_{i=1}^ni^3=\frac{n^2(n+1)^2}{4}



To prove the basis, I substitute 1 for n in S(n):




\sum_{i=1}^11^3=1=\frac{1^2(2)^2}{4}



Great. For the inductive step, I assume S(n) to be true and prove S(n+1):



\sum_{i=1}^{n+1}i^3=\frac{(n+1)^2(n+2)^2}{4}



Considering the sum on the left side:



\sum_{i=1}^{n+1}i^3=\sum_{i=1}^ni^3+(n+1)^3




I make use of S(n) by substituting its right side for \sum_{i=1}^ni^3:



\sum_{i=1}^{n+1}i^3=\frac{n^2(n+1)^2}{4}+(n+1)^3



This is where I get a little lost. I think I expand the equation to be



=\frac{(n^4+2n^3+n^2)}{4}+(n+1)^3



but I'm not totally confident about that. Can anyone provide some guidance?

No comments:

Post a Comment

analysis - Injection, making bijection

I have injection f \colon A \rightarrow B and I want to get bijection. Can I just resting codomain to f(A)? I know that every function i...