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Differentiation can be used to find the sum of \( n \) terms or infinite terms in certain cases. e.g. we know that for \( |x|1 \)
\[
\begin{aligned}
\frac{1}{1-x} & =1+x+x^{2}+\ldots \\
\frac{1}{(1-x)^{2}} & =1+2 x+3 x^{2}+4 x^{3}+\ldots+n x^{n-1}+\ldots \\
& =\sum_{n=1}^{\infty} n x^{n-1} \\
\frac{2}{(1-x)^{3}} & =\sum_{n=2}^{\infty} n(n-1) x^{n-2}
\end{aligned}
\]
\( 2\left[(n-1)^{n} C_{1} x^{n-2}+(n-3){ }^{n} C_{3} x^{n-4}+(n-5){ }^{n} C_{5}\right. \)
\( x^{n-6}+\ldots n \) terms \( ] \) is equal to
(a) \( n\left[(x+1)^{n-1}+(x-1)^{n-1}\right] \)
(b) \( n\left[(x+1)^{n-1}-(x-1)^{n-1}\right] \)
(c) \( 2 n(x+1)^{n+1} \)
(d) \( n(x+1)^{n-1} \)
\( \mathrm{P} \)
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