Prove that $(a^n - b^n) = (a-b) \sum_{i=1}^n a^{i-1} b^{n-i}$

Question

Let it be $a, b \in\Bbb R$. Prove that $\forall n \in\Bbb N$, $(a^n - b^n) = (a-b) \sum_{i=1}^n a^{i-1} b^{n-i}$.

Deduce the formula of the geometric sum: $\forall a ≠ 1, \sum_{i=0}^n a^i = \frac{a^{n+1}-1}{a-1} $.

For the first part I tried making a proof by induction, but I got stuck when checking for P(n+1)... for the second part I haven't started but I would appreciate any hint.

Thanks!

Answer 1

Thanks to @drhab & @RRL

Indeed

$f(n) - f(0) = a^n b^{n-n} - a^0 b^{n-0} = a^n - b^n $

As for the second part, the formula of the geometric sum:

$S = \sum_{i=0}^n a^i = 1 + a + a^2 + ... + a^n$

If i take the sum and multiply it times $a$ I get:

$S\times a = a + a^2 + a^3 + ... + a^{n+1}$

Then I sustract $S$ from $S \times a$ and solve for $S$:

$(S \times a) - S = a^{n+1} - 1$

$S (a - 1) = a^{n+1} - 1$

$S = \frac{a^{n+1} - 1} {a-1}$

Answer 2

For the first part you can indeed use induction, but you can also do without: $$\left(a-b\right)\sum_{i=1}^{n}a^{i-1}b^{n-i}=\sum_{i=1}^{n}a^{i}b^{n-i}-\sum_{i=1}^{n}a^{i-1}b^{n+1-i}=\cdots$$

Just work this out.