<s>[INST] Proof via induction $1\cdot3 + 2\cdot4 + 3\cdot5 + \cdots + n(n+2) = \frac{n(n+1)(2n+7)}{6}$ (b) Prove that for every integer $n \ge 1$, $$1\cdot3 + 2\cdot4 + 3\cdot5 + \cdots + n(n+2) = \frac{n(n+1)(2n+7)}{6}$$
This is the second part of a two part question.  Part (a) was the following: 
Write the sum: $1\cdot3 + 2\cdot4 + 3\cdot5 + \cdots + n(n+2)$ using summation notation. 
It was simple enough :  $\sum  k(k+2)$.
For this question, the base case $(n=1)$ holds, as $1\cdot(1+2) = 3 = (1\cdot2\cdot9)/6$.
Induction step: Assume the above holds for all $n = k$, prove that it holds for all $n = k+1$
I'm a bit lost from here, help?
 [/INST] Note that
$$
k(k + 2) = k^2 + 2k = k^2 - k + 3k = 2{k \choose 2} + 3{k \choose 1}
$$
Definition 1: Let $\{x_n\}$ a sequence. $\Delta x_n = x_{n+1} - x_n$.
Definition 2: If $\Delta X_n = x_n$, then $X_n = \Delta^{-1}x_n + C$ with $C \in \mathbb{R}$.
Proposition 1: $\Delta^{-1}$ is linear and
$$
\Delta \biggl(\sum_{k =n_0}^{n-1}x_k \biggr) = x_n \quad \text{and} \quad \Delta {n \choose p} = {n \choose p - 1}
$$
Proof. Exercise.
Proposition 2: $\sum_{k =m}^{n-1}x_k = \Delta^{-1}x_n\biggl|_{m}^{n} = X_n - X_m$.
Proof. Exercise.
Proposition 3: $\Delta^{-1}{n \choose p} = {n \choose p+1} + C$.
Proof. Exercise.
Thus,
$$
\sum_{k=1}^{n}k(k+2) = \sum_{k=1}^{n}\biggl[2{k \choose 2} + 3{k \choose 1}\biggr]
= \biggl[2{k \choose 3} + 3{k \choose 2}\biggr]_{1}^{n+1} 
$$
$$
= 2{n+1 \choose 3} + 3{n+1 \choose 2} = \dfrac{1}{3}(n+1)n(n-1) + \dfrac{3}{2}(n+1)n
$$
$$
=n(n+1)\biggl[\dfrac{n}{3} - \dfrac{1}{3} + \dfrac{3}{2}\biggr] = \dfrac{n(n+1)(2n+7)}{6}
$$
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