<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] Consider the telescoping sum
$$\sum_{k=0}^n(k+1)^4-k^4=n^4+4n^3+6n^2+4n.$$
We will work with the left-hand side and rewrite it as
$$\sum_{k=0}^n (k+1)^4-k^4=\sum_{k=0}^n 4k^3+6k^2+4k+1.$$
Now we have $$4\sum_{k=0}^n k^3+6\sum_{k=o}^n k^2+4\sum_{k=0}^n k+\sum_{k=0}^n 1.$$ We can let $4\sum_{k=0}^n k^3=4S$ since we are trying to find what S equals. Now we have
$$4S+{6n(n+1)(2n+1)\over6}+4{n(n+1)\over2}+n.$$
Which becomes $$4S+2n^3+5n^2+4n.$$
Thus we have $$n^4+4n^3+6n^2+4n=4S+2n^3+5n^2+4n,$$
and after a little algebra we will obtain $$S=n^2(n+1)^2\over4.$$
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