Showing a compact form of a summation.

Question

How can I show that:

$$P^k \left(1 + \frac {2}{p} + \frac {3}{p^2} + \frac {4}{p^3} + ..... + \frac{k}{p^{k-1}} + \frac{k+1}{p^k}\right) = \sum_{i = 0}^{k} (k + 1 -i) p^i$$

Could anyone explain this for me please?

Answer 1

It is just a change of summation index: $$p^k\sum_{l=0}^k\frac{l+1}{p^l}= \sum_{l=0}^k(l+1)p^{k-l} \stackrel{i=k-l}{=}\sum_{i=k}^0(k-i+1)p^{i} = \sum_{i=0}^k(k-i+1)p^{i}$$