Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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


share|cite|improve this question
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
up vote 3 down vote accepted

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






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 $n$".

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.