To prove:
$n+2(n-1)+3(n-2)+...+2(n-1)+n={n(n+1)(n+2)\over 6}$
I assume the statement was derived from:
$n+2(n-1)+3(n-2)+...+(n-1)(n-(n-2))+n(n-(n-1))={n(n+1)(n+2)\over 6}$
For $n=k+1$
$(k+1)+2(k)+3(k-1)+...+2(k)+(k+1)={(k+1)(k+2)(k+3)\over 6}$
How do you use substitution from $n=k$, to prove $n=k+1$ when there are no duplicate terms?
Hint: You can just extract $1$ from each product to use the induction hypothesis:
$$ \begin{aligned} \phantom{=}&1\cdot(k+1) + 2\cdot k + 3\cdot(k-1) + \cdots + k\cdot 2 + (k+1) \\ =&(1\cdot k + \color{red}{1}) + (2\cdot (k-1) + \color{red}{2}) + \cdots + (k\cdot 1 + \color{red}{k}) + \color{red}{k+1}\\ =&\frac{(k-1)k(k+1)}6 + \color{red}{\frac{(k+1)(k+2)}2} \end{aligned} $$
