Help with understand the following inequality steps.

Question

Given the following inequality:

$$ \begin{align} n&\ge1+(t-1)\sum_{i=1}^h 2t^{i-1}\\ &=1+2(t-1)\left(\frac{t^h-1}{t-1}\right)\\ &=2t^h-1 \end{align} $$

It's not clear to me how the first line gets simplified to the second, mainly the $\left(\frac{t^h-1}{t-1}\right)$ part?

Answer 1

Let $t \not =1;$

$S_m:= \sum_{i=1}^{m}t^{i-1}=$

$= 1 + t+t^2+.........t^{m-1};$

$tS_m= t+t^2+.......+t^{m-1} + t^m.$

Hence :

$tS_m- S_m =(t-1)S_m= t^m-1;$

$S_m = \dfrac{t^m-1}{t-1}.$