Notation Question. How to approach case where n = 1 if pattern starts at 2?

Question

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!

The formula is Ok for $n=2$. For $n=1$, the part $(2 +3+\ldots + n)$ will vanish and we are left with: $1+(1+1)=3$. But $(1+1)^2-1=4-1=3$ and it's fine. — Mauro ALLEGRANZA, Feb 13 '18 at 07:16
That's the problem when using $\ldots$ in a formula - there's always some doubt left. But here it seems safe to assume that $(2+3+\ldots+n)$ is supposed to be the sum of the $n-1$ integers from $2$ to $n$ inclusive. For $n=1$, this means we have to sum zero integers, so the sum is $0$ (without worrying which integers) — Hagen von Eitzen, Feb 13 '18 at 07:22

score 1 · Answer 1 · answered Feb 13 '18 at 15:40

1

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 $\sum_{i=2}^1i=0$.

answered Feb 13 '18 at 15:40

1 Answers1