What is $(1+x+\cdots+x^{n-1})^2$?

Question

This may seem like a pretty basic question, but I am struggling to find a nice way of expressing $(1+x+\cdots+x^{n-1})^2$ where $n\in\mathbb{Z}_{\geq 2}$. Plugging in some numbers yields a clear pattern,

$(1+x+\cdots+x^{n-1})^2=1+2x+\cdots+nx^n+\cdots+2x^{2n-3}+x^{2n-2}.$

However, I can't seem to find a formal expression of this, or a proof. Can anyone point me in the right direction?

Answer 1

So you can notice that there is a very nice elementary formula for partial sum of geometric progression, i.e. $$ \sum_{i=0}^{n-1} x^i = \frac{1-x^n}{1-x} $$ It comes from the following argument $$ \sum_{i=0}^{n-1} x^i (1-x) = \sum_{i=0}^{n-1} x^i - \sum_{i=0}^{n-1} x^{i+1} = 1 - x^{n-1+1} = 1-x^n $$ Therefore you can square to obtain $$ \left( \sum_{i=0}^{n-1} x^i \right)^2 = \left( \frac{1-x^n}{1-x} \right)^2 = \frac{1+x^{2n} - 2x^n}{1+x^2-2x} $$

Answer 2

$$1+x+...+x^{n-1} = \frac{1\cdot(1-x^n)}{1-x},\quad x\neq 1$$

Then $$1+x+...+x^{n-1} = \bigg(\frac{1-x^n}{1-x}\bigg)^2=\frac{1+x^{2n}-2x^n}{1+x^2-2x}$$

Answer 3

Its a sum of the GP with common ratio x. Thus apply the sum for a GP and then square it.