<s>[INST] Evaluate the sum of cubes Evaluate $1^3 + 2^3 + 3^3 + . . . + n^3.$
Can I get a hint? I'm really stuck and don't know how to break this problem down.
 [/INST] We'll prove the following using induction:
$$1^3 + 2^3 + 3^3 + 4^3 + 5^3+...+n^3 = [S(n)]^2$$
Where $S(n)$ is the sum of the consecutive numbers from $1$ to $n$. We know that $S(n) = \frac{n(n+1)}{2}$. So we have:
$$1^3 + 2^3 + 3^3 + 4^3 + 5^3+...+n^3 = \left(\frac{n(n+1)}{2}\right)^2$$
Now let's prove it:
Basis $n=1$
$$1^3 = \left(\frac{1(1+1)}{2}\right)^2$$
$$1 = 1$$
Which is true:
Hypothesis $n=k$ 
Assume that the follwoing holds for some number $k$:
$$1^3 + 2^3 + 3^3 + 4^3 + 5^3+...+k^3 = \left(\frac{k(k+1)}{2}\right)^2$$
Unductive step $n=k+1$
We'll prove that it holds for every number $k+1$ so we have:
$$\left(\frac{(k+1)(k+2)}{2}\right)^2 = \left(\frac{(k+1)k+2(k+1)}{2}\right)^2 = \left(\frac{k(k+1)}{2}\right)^2 + \frac{4k(k+1)(k+1)}{4} + \left(\frac{2(k+1)}{2}\right)^2$$
Substitute for the first term from the hypothesis and expand the other two. and we have:
$$\left(\frac{(k+1)(k+2)}{2}\right)^2 = 1^3 + 2^3 + 3^3 + 4^3 + 5^3+...+k^3 + k(k+1)(k+1) + (k+1)^2$$
$$\left(\frac{(k+1)(k+2)}{2}\right)^2 = 1^3 + 2^3 + 3^3 + 4^3 + 5^3+...+k^3 + (k+1)^2(k + 1) = 1^3 + 2^3 + 3^3 + 4^3 + 5^3+...+k^3 + (k+1)^3$$
Q.E.D.
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