Inductive proof and summation: $\sum\limits_{i=1}^n i^3 = \frac{n^2(n+1)^2}{4}$

Question

The problem asks me to prove by induction that: $$\sum_{i=1}^n i^3 = \frac{n^2(n+1)^2}{4}$$

I've worked through it at least half a dozen times, checked my math fastidiously, can't seem to figure it out, so here I am.

For a base case I use 2, gets me 9 for both, inductive step leaves me with: $$\sum_{i=1}^{k+1} i^3 = \frac{(k+1)^2 * (k+2)^2}{4}$$

Expanding things, I square the binomials to get $(k^2+2k+2)(k^2+4k+4)$ Taking it from there (and I won't go through the steps of multiplying them together and combining like terms) I am left with: $$\frac{k^4+6k^3+14k^2+16k+8}{4}$$ Plugging in 2 again yields 36 for the summation portion but 40 for the fraction part, trying it with 3 yields equally unequal answers. If someone could please point me to the spot where I am making my error it would be greatly appreciated, thanks in advance!

Answer 1

You've gone wrong on the easy bit: $(k+1)^2=k^2+2k+1$!

---

You can avoid full expansion of the numerator in various ways, for example by computing

$$\frac {(n+1)^2(n+2)^2}4-\frac{n^2(n+1)^2}4=\frac {(n+1)^2}4\cdot\left ((n+2)^2-n^2\right)$$

And the final bracket is the difference of two squares and comes to $2(2n+2)=4(n+1)$

Answer 2

The inductive step would require showing that $$ \left[n^{2}(n+1)^{2}/4\right]+(n+1)^{3}=(n+1)^{2}((n+1)+1)^{2}/4. $$