<s>[INST] Summation and proof by induction question: $\sum_{j=1}^{n}j(j+1)(j+2)=\frac{n(n+1)(n+2)(n+3)}{4}$ 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?
 [/INST] We want to show that

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

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
$$\sum_{j=1}^{n}j(j+1)(j+2)=\frac{n(n+1)(n+2)(n+3)}{4}$$
and wish to prove
$$\sum_{j=1}^{n+1}j(j+1)(j+2)=\frac{(n+1)(n+2)(n+3)(n+4)}{4}$$
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.
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