# $k(k^n - (k-1)^n) = k^{n+1}-(k-1)^{n+1}-(k-1)^n$

In Feller's Intro to Probability Theory Vol. 1. there is a step I don't know how the author proceded.

You have the full source here

In first place we have (you can ignore the $$N^n$$):

$$N^n p_k = k^n - (k-1)^n$$

Now we have:

$$E(X) = \sum_{k = 1}^Nkp_k$$ $$= N^{-n}\sum_{k = 1}^N \{k (k^n - (k-1)^n)\} = N^{-n}\sum_{k = 1}^N \{k^{n+1}-(k-1)^{n+1}-(k-1)^n\}$$

Which is followed by the question:

$$k(k-1)^n \overset{?}{=} (k-1)^{n+1}+(k-1)^n$$

This algebra is the argument for an approximation he later does on the same page number, however I don't believe the last steps he made are correct.

• Have you tried factoring $(k-1)^n$ out? Commented Apr 16, 2021 at 4:19
• I've expanded the expression, but I get a monster sumatory of terms. I'm stuck in just that step. Commented Apr 16, 2021 at 4:21
• No. It is very simple to show $k(k-1)^n = (k-1)^{n+1}+(k-1)^n$. Commented Apr 16, 2021 at 4:22
• How? I just can't see it yet. Commented Apr 16, 2021 at 4:24
• Here's a hint: $1-1=0$, and equivalently $x+1-1=x$ Commented Apr 16, 2021 at 4:25

$$k(k-1)^n - (k-1)^n = (k-1)(k-1)^n = (k-1)^{n+1}$$
and add $$(k-1)^n$$ to both sides.
Don't try expanding the expression out; try factoring the right-hand side: $$(k - 1)^{n+1} = (k-1)^n(k-1)$$ so that $$(k - 1)^{n + 1} + (k - 1)^{n} = (k-1)^{n}(k - 1) + (k - 1)^{n} = \left[(k - 1) + 1\right](k-1)^{n} = k(k-1)^{n}.$$