# 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.

• Oct 24 '13 at 17:59
• Jul 4 '14 at 17:27

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.

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 $$S=n^2(n+1)^2\over4.$$

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.