<s>[INST] Discrete math induction problem. I am stuck at this step in the inductive process and I was wondering if someone can help me out from where I am stuck.
Question:
if $n$ is a positive integer, prove that,
$$1\cdot2+2\cdot3+3\cdot4+\cdots+n\cdot(n+1) = \frac{n(n+1)(n+2)}3$$
Basic step is $1$, then assume $k$ is true and we get,
$$1\cdot2+2\cdot3+3\cdot4+\cdots+k\cdot(k+1) = \frac{k(k+1)(k+2)}3$$
After that I am stuck because the answer key is telling me that the next step is to add $(k+1)(k+2)$, but I don't know why we do that. Aren't we supposed to just plugin $k+1$ into all $k$ variables. Why are we adding $(k+1)(k+2)$ to both sides?
 [/INST] Finishing up the induction proof,\begin{align}
1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + \cdots + (k+1)(k+2) &= [1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + \cdots + k(k+1)]+(k+1)(k+2) \\
&= \frac{k(k+1)(k+2)}{3}+(k+1)(k+2) \\
&= \frac{k(k+1)(k+2)}{3}+\frac{3(k+1)(k+2)}{3} \\
&= \frac{k(k+1)(k+2)+3(k+1)(k+2)}{3} \\
&= \frac{(k+1)(k+2)(k+3)}{3} 
\end{align}
where the last equality is by hypothesis. Now simplify the last expression accordingly and you will get your result.
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