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enter image description hereI have been stuck on this example on Trench introduction to real analysis.

I don't understand how the sum expression came.

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    See [here](http://math.stackexchange.com/questions/2031059/prove-that-an-bn-a-ban-1-an-2b-cdots-bn-1/2031074#2031074) – 2017-02-03
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    Use $$a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b^1+a^{n-3}b^2+...+a^1b^{n-2}+b^{n-1})$$ – 2017-02-03
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    I up-voted the question, and its total is now $0$, so somewhat down-voted it. If there's an objection to it, it cannot lead to any improvement until the nature of the objection is known. – 2017-02-03

2 Answers 2

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\begin{align} \frac{x^5 - x_0^5}{x-x_0} & = \frac{(x-x_0)(x^4 + x^3x_0 + x^2 x_0^2 + x x_0^3 + x_0^4)}{x-x_0} \\[10pt] & = \frac{x-x_0}{x-x_0} (x^4 + x^3x_0 + x^2 x_0^2 + x x_0^3 + x_0^4) \\[10pt] & = \phantom{\frac{x-x_0}{x-x_0} (} x^4 + x^3x_0 + x^2 x_0^2 + x x_0^3 + x_0^4 \\[10pt] & = \sum_{k=0}^4 x^{5-k-1} x_0^k \end{align} (And as with $5$, so also with other values of $n$.)

As $x\to x_0,$ the sum approaches $\displaystyle \sum_{k=0}^4 x_0^{5-k-1} x_0^k = \sum_{k=0}^4 x_0^{5-1} = 5x_0^{5-1}.$

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    Someone down-voted this, but that cannot lead to any improvement unless the objection is explained. – 2017-02-03
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$$f'(x_0)=\lim_{x \rightarrow x_0}\dfrac{x^n-x_0^n}{x-x_0}=\\ \lim_{x \rightarrow x_0}\dfrac{(x-x_0)(x^{n-1}+x^{n-2}x_0^1+x^{n-3}x_0^3+\cdots+x^{1}x_0^{n-2}+x_0^{n-1})}{x-x_0}=\\ \lim_{x \rightarrow x_0}\dfrac{(x^{n-1}+x^{n-2}x_0^1+x^{n-3}x_0^3+\cdots+x^{1}x_0^{n-2}+x_0^{n-1})}{1}=\\(x_0^{n-1}+x_0^{n-2}x_0^1+x_0^{n-3}x_0^3+\cdots+x_0^{1}x_0^{n-2}+x_0^{n-1})=nx_0^{n-1}$$and twice looking $$\lim_{x \rightarrow x_0}\dfrac{(x-x_0)(x^{n-1}+x^{n-2}x_0^1+x^{n-3}x_0^3+\cdots+x^{1}x_0^{n-2}+x_0^{n-1})}{x-x_0}=\\ \lim_{x \rightarrow x_0}\dfrac{(x-x_0)}{x-x_0}(x^{n-1}+x^{n-2}x_0^1+x^{n-3}x_0^3+\cdots+x^{1}x_0^{n-2}+x_0^{n-1})=\\\lim_{x \rightarrow x_0}\dfrac{(x-x_0)}{x-x_0}\sum_{k=0}^{n-1}x^{\color{red}{n-1-k}}x_0^{k}$$