Understanding $\frac {b^{n+1}-a^{n+1}}{b-a} = \sum_{i=0}^{n}a^ib^{n-i}$

Question

I'm going through a book about algorithms and I encounter this.

$$\frac {b^{n+1}-a^{n+1}}{b-a} = \sum_{i=0}^{n}a^ib^{n-i}$$

How is this equation formed? If a theorem has been applied, what theorem is it?

[Pardon me for asking such a simple question. I'm not very good at maths.]

Answer 1

Multiply both sides by b-a, watch for the cancling of terms, and you will have your answer.

Answer 2

Send $i \mapsto i-1$ then $ \displaystyle f(n) = \sum_{i=0}^{n}a^{i}b^{n-i} = \sum_{i=1}^{n+1}a^{i-1}b^{n-i+1}\implies a f(n) = b\sum_{i=1}^{n+1}a^{i}b^{n-i} $

$\displaystyle \implies af(n) = -b^{n+1}+a^{n+1}+b\sum_{i=0}^{n}a^{i}b^{n-i} =-b^{n+1}+a^{n+1}+bf(n) $ and rearranging:

$\displaystyle bf(n)-af(n) = b^{n+1}-a^{n+1} \implies (b-a)f(n) = b^{n+1}-a^{n+1} \implies f(n) = \frac{b^{n+1}-a^{n+1}}{b-a} $.