I can't figure this out based on examples in textbooks, etc.

Show via induction that $\sum_{j=1}^{n}j(j+1)(j+2)=\frac{n(n+1)(n+2)(n+3)}{4}$

So far, I have:

(a) base case

$P(1)= 1(1+1)(1+2) = \frac{1(1+1)(1+2)(1+3)}{4} = 6 = 6$

(b) inductive step $P(k)\rightarrow P(k+1)$

If $P(k)$ is true, then

$P(k+1) = (2)(3)+(2)(3)(4)+(3)(4)(5)+\cdots+(k+1)(k+2)(k+3)= \frac{(k+1)(k+2)(k+3)(k+4)}{4}$

But this doesn't seem to be going anywhere. Any ideas?

  • $\begingroup$ No problem at all. :) $\endgroup$ Jul 29, 2013 at 22:18

3 Answers 3


We want to show that


Base case, $n=0$. It is easy to see that the left-hand-side and right-hand-side are both equal to $0$, so the equality is true. You can also let the base case be $n=1$, and then both sides are equal to $6$.

Induction step. We are given that


and wish to prove


So we write

$$\sum_{j=1}^{n+1}j(j+1)(j+2) = \left(\sum_{j=1}^{n}j(j+1)(j+2)\right) + (n+1)(n+2)(n+3)$$

By assumption, this is equal to

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

Factoring, this becomes

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

And the proof is complete.

  • $\begingroup$ Great explanation, thanks! $\endgroup$
    – MNRC
    Jul 29, 2013 at 22:25

If $$\sum_{j=1}^{n}j(j+1)(j+2)=\frac{n(n+1)(n+2)(n+3)}{4}$$ then $$\sum_{j=1}^{n+1}j(j+1)(j+2)=\sum_{j=1}^{n}j(j+1)(j+2)+(n+1)(n+2)(n+3)$$ $$=\frac{n(n+1)(n+2)(n+3)}{4}+(n+1)(n+2)(n+3)=$$ $$=\frac{n(n+1)(n+2)(n+3)}{4}+\frac{4(n+1)(n+2)(n+3)}{4}=$$ $$=\frac{(n+1)(n+2)(n+3)(n+4)}{4}$$


First, to improve the logic of your proof, you don't want to put the thing that you are trying to prove, $P(k+1) = (2)(3)+(2)(3)(4)+(3)(4)(5)+\cdots+(k+1)(k+2)(k+3)= \frac{(k+1)(k+2)(k+3)(k+4)}{4}$ at the beginning of (the second part of) your proof. Same goes for the base case where you you are trying to show that the left and right side both equal 6, but you are doing them in parallel. Try either:

$P(1)= 1(1+1)(1+2) = 6 = \frac{1(1+1)(1+2)(1+3)}{4}$


$P(1)= 1(1+1)(1+2) = 6$


$\frac{1(1+1)(1+2)(1+3)}{4} = 6$

Now for the actual proof:

We have $\sum_{j=1}^{n}j(j+1)(j+2)=\frac{n(n+1)(n+2)(n+3)}{4}$. Now take $\sum_{j=1}^{n+1}j(j+1)(j+2)=\sum_{j=1}^{n}j(j+1)(j+2) + (n+1)(n+2)(n+3)$

and by the inductive hypothesis

$\sum_{j=1}^{n}j(j+1)(j+2) + (n+1)(n+2)(n+3) = \frac{n(n+1)(n+2)(n+3)}{4} + (n+1)(n+2)(n+3) = \frac{n(n+1)(n+2)(n+3)+4(n+1)(n+2)(n+3)}{4}=\frac{(n+4)(n+1)(n+2)(n+3)}{4}$ which concludes our proof.

It is generally not a good idea to expand your sums like this $(2)(3)+(2)(3)(4)+(3)(4)(5)+\cdots+(k+1)(k+2)(k+3)$. It is sometimes helpful for spotting patterns like telescoping sums, but it generally turns into a mess.


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