For purposes of induction, I'm given to prove
$1 + 2(2+3+...+n) + (n+1) = (n+1)^2 - 1$, for all $n$ in natural numbers (defined to start from 1).
I want to do the base case of induction using the natural number 1 for $n$, but I'm unsure how to approach the portion $(2+3+...+n)$ when $n = 1$.
Thank you for the help!
Perhaps
$$1 + 2(2+3+\cdots+n) + (n+1) = (n+1)^2 - 1$$
is intended to be equivalent to
$$1 + \left(2\sum_{i=2}^ni\right) + (n+1) = (n+1)^2 -1$$
You can simplify this further, but by the property of an empty sum, if $n=1$ then $\sum_{i=2}^1i=0$.