Friday, April 19, 2019

algebra precalculus - I'm having trouble with induction. Prove $1 + 2^3 + 3^3 + ... + n^3 = frac{((n^2)(n+1)^2)}4$





I started a new course and I'm expected to know this stuff, and I'm having trouble learning some on my own.
I'm stuck with this problem:
Prove $1 + 2^3 + 3^3 + ... + n^3 = \frac{((n^2)(n+1)^2)}4$ using induction (the $1+2^3...$ is written in sum notation, although I don't know how to enter that here, sorry).
I started substituting $n \rightarrow n+1$ but I don't know what to do next. Any help would be extremely appreciated.


Answer



$$1+2^3+...+n^3+(n+1)^3\underset{Hyp.}{=}\frac{n^2(n+1)^2}{4}+(n+1)^3=\frac{n^2(n+1)^2+4(n+1)^3}{4}=\frac{(n+1)^2(n^2+4n+4)}{4}=\frac{(n+1)^2(n+2)^2}{4}.$$


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...