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.
 A: Hint:
The sum of first powers to n is quadratic, and of 2nd powers cubic, so try writing down a general quartic. Find the first few sums and solve the resulting system of linear equations. Then prove your result by induction. 
Alternatively, note that $(k+1)^4-k^4=4k^3+6k^2+4k+1$ so the sum of both $k$ from 1 to $n$ is the same. Then see how it cancels down nicely.
A: 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 $$S=n^2(n+1)^2\over4.$$
A: 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.
