<s>[INST] 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 
$$\frac{(k+1)(k+2)(k+3)}{3}$$
 [/INST] 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 $n$".
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