# 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

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

-
Factor $(k+1)(k+2)$ out from the expression after "Which is equal to". – David Mitra Feb 2 '13 at 13:11
Note, if you wanted to subvert the problem stated, you could perform induction separately on $\sum n^2$ and $\sum n$. – half-integer fan Feb 2 '13 at 13:20
I don't understand, @Andrew: you asked a very, very similar questions some hours ago and you had exactly the same algebraic problem as you have here (non-factoring when possible)...are you really making an efforto to learn?! – DonAntonio Feb 2 '13 at 16:25
@Andrew : Please. Don't promiscuously interchage lower-case $n$ with capital $N$, nor lower-case $k$ with capital $K$, as if they were the same thing. Mathematical notation is case-sensitive. To anyone reading what you write who knows standard conventions it will look as if you can't spell. – Michael Hardy Feb 2 '13 at 16:54

## 1 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 $n$".

-
I know That the Final answer is $$\frac{(k+1)(k+2)(k+3)}{3}$$ by adding a k + 1 to every unknown from the Step 2 Which is _Assume N = K_$$\frac{(k)(k+1)(k+2)}{3}$$ – Andrew Feb 2 '13 at 13:19
Thanks for the Quick Edit, I only have one more question, What I'm Not understanding is how $$\frac{k(k+1)(k+2)}{3}$$ was changed to $$\frac{k}{3}+1$$ – Andrew Feb 2 '13 at 13:26