discrete mathematics - Proving $(1)(2) + (2)(3) + (3)(4) + cdots+ (n) (n+1) = frac{(n)(n+1)(n+2)}{3}$ by induction

I have the Following Proof By Induction Question:

$$(1)(2) + (2)(3) + (3)(4) + \cdots+ (n) (n+1) = \frac{(n)(n+1)(n+2)}{3}$$

Can Anybody Tell Me What I'm Missing.

This is where I've Gone So Far.

---

Show Truth for N = 1

LHS = (1) (2) = 2

RHS = $$\frac{(1)(1+1)(1+2)}{3}$$

Which is Equal to 2

Assume N = K

$$(1)(2) + (2)(3) + (3)(4) + \cdots+ (k) (k+1) = \frac{(k)(k+1)(k+2)}{3}$$

Proof that the equation is true for N = K + 1

$$(1)(2) + (2)(3) + (3)(4) + \cdots+ (k) (k+1) + (k+1) (k + 2)$$

Which is Equal To:
$$\frac{(k)(k+1)(k+2)}{3} + (k+1) (k + 2)$$

This is where I've went so far

If I did the calculation right the Answer should be

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

Answer

Your proof is fine, but you should show clearly how you got to the last expression.

$\dfrac{k(k+1)(k+2)}{3}+(k+1)(k+2)$

$=\dfrac{k}{3}(k+1)(k+2)+(k+1)(k+2)$

$=(\dfrac{k}{3}+1)(k+1)(k+2)$

$=\dfrac{k+3}{3}(k+1)(k+2)$

$=\dfrac{(k+1)(k+2)(k+3)}{3}$.

You should also word your proof clearly. For example, you can say "Let $P(n)$ be the statement ... $P(1)$ is true ... Assume $P(k)$ is true for some positive integer $k$ ... then $P(k+1)$ is true ... hence $P(n)$ is true for all positive integers $n$".