Help needed in understanding the telescoping sum for series $(a-b)\sum_{i=0}^{n-1}a^ib^{n-1-i}$.

Question

Request help in understanding the telescoping sum for the given series.
For $n\geq1$, \begin{align} (a-b)\sum_{i=0}^{n-1}a^ib^{n-1-i}&=\sum_{i=0}^{n-1}a^{i+1}b^{n-1-i}-\sum_{i=0}^{n-1}a^ib^{n-i}\\ &=\sum_{i=0}^{n-1}(a^{i+1}b^{n-(i+1)}-a^ib^{n-i})\\ &=a^n-b^n&&(\text{telescoping sum}) \end{align}

I mean the conversion from the second last step to the last step is not clear.

Answer 1

Given$$ \sum_{i=0}^{n-1}(a^{i+1}b^{n-(i+1)}-a^ib^{n-i})\\$$ Change the index of summation on the first sum: $$=\sum_{i=1}^na^ib^{n-i} - \sum_{i=0}^{n-1}a^ib^{n-i}$$ Take out the $i=n$ term from the first sum and the $i=0$ term from the second, combining the rest:

$$=a^n+\sum_{i=1}^{n-1}(a^ib^{n-i}-a^ib^{n-i})-b^n$$

$$=a^n+0-b^n = a^n-b^n$$

Answer 2

If you write $c_i = a^ib^{n-i}$, then your sum turns into $$\sum_{i=0}^{n-1}(a^{i+1}b^{n-(i+1)}-a^ib^{n-i}) = \sum_{i=0}^{n-1}(c_{i+1}-c_i) = c_n - c_0 = a^n-b^n$$