<s>[INST] Prove that $1^3 + 2^3 + 3^3 +\cdots+ n^3 = \frac14n^4 + \frac12n^3 + \frac14n^2$ I have to prove that this is true using mathematical induction. 
I have this: 
for every $n \in \mathbb N$: $1^3 + 2^3 + 3^3 + ... + n^3 = \frac 14n^4 + \frac 12n^3 + \frac 14n^2$
for $n = 1: 1^3 = 1/4 + 1/2 + 1/4$, hence $P(1)$ is true.
Let $N \in \mathbb N$ be given and assume that $P(N)$ is true, that is $$1^3 + 2^3 + 3^3 + ... + N^3 = \frac 14N^4 + \frac 12N^3 + \frac 14N^2$$
For n = $N$ + 1: 
And now what? I just couldn't solve it.
 [/INST] Let $P(N)$ is true, that is $1^3+\ldots+n^3=\frac{1}{4}n^4+\frac{1}{2}n^3+\frac{1}{4}n^2$
Then for $P(N+1)$, $1^3+\ldots+n^3+(n+1)^3=\frac{1}{4}n^4+\frac{1}{2}n^3+\frac{1}{4}n^2+n^3+3n^2+3n+1$
$=\frac{1}{4}n^4+n^3+\frac{3}{2}n^2+n+\frac{1}{4}+\frac{1}{2}n^3+\frac{3}{2}n^2+\frac{3}{2}n+\frac{1}{2}+\frac{1}{4}n^2+\frac{1}{2}n+\frac{1}{4}$
$=\frac{1}{4}(n+1)^4+\frac{1}{2}(n+1)^3+\frac{1}{4}(n+1)^2$
That is $P(N+1)$ is true.
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