<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] It might also be helpful to go backwards to see that it makes sense. 
Specifically, 
$$S_{n+1} - S_n = \frac{(n+2)(n+1)(n)}{3} - \frac{(n+1)(n)(n-2)}{3} = \frac{1}{3}\left(n^3 + 3n^2 + 2n - n^3 + n\right) = n(n+1)$$
Going backwards, you can more easily see that $$S_{n+1} = S_n + n(n+1)$$
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