Mathematical induction proof problem: $\sum_{i=1}^{n-1} i(i+1) = \frac{n(n+1)(n-1)}3$

Question

I am having difficulty proving the inductive hypothesis $(k+1)$ for the following statement: $$\sum_{i=1}^{n-1} (i(i+1)) = \frac{(n)(n+1)(n-1)}{3}$$

This is what I have so far:

$$(Step \ 1) \sum_{i=1}^{k-1} (i(i+1)) + (k+1)(k+2) = \frac{(k+1)(k+2)(k)}{3}$$

$$(Step \ 2)\frac{(k)(k+1)(k-1)}{3} + (k+1)(k+2) = \frac{(k+1)(k+2)(k)}{3}$$

$$(Step \ 3) \frac{(k^3 + 3k^2 +8k + 6)}{3} \ != \frac{(k^3 + 3k^2 + 2k)}{3}$$

I do not know what I did wrong. Thank you.

Answer 1

Inductive hypothesis (=the claim holds for $k$): $$\sum_{i=1}^{k-1} i(i+1) = \frac{k(k+1)(k-1)}3.$$

Using this you want to show that it is true for $k+1$ instead of $k$: $$\sum_{i=1}^{k} i(i+1) = \frac{(k+1)(k+2)k}3.$$ (This is the inductive step.)

You have $$\sum_{i=1}^{k} i(i+1) = \sum_{i=1}^{k-1} i(i+1) + k(k+1) \overset{(*)}= \frac{k(k+1)(k-1)}3 + k(k+1) = k(k+1) \left(\frac{k-1}3 +1\right) = k(k+1)\cdot \frac{k+2}3 = \frac{k(k+1)(k+2)}3. $$

The step marked $(*)$ is the step where we are using the inductive hypothesis. Notice that in the inductive step I was adding $k(k+1)$. You were attempting a proof along the same lines, but you were adding $(k+1)(k+2)$.