Thursday, January 17, 2019

algebra precalculus - Prove by induction $sum_{i=1}^ni^3=frac{n^2(n+1)^2}{4}$ 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...