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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


here pag. 226


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.

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

2 Answers 2

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

and add $(k-1)^n$ to both sides.

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

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