Question

The nth term in the Maclaurin's expansion of 1/1-x
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Answer 1

$\textbf{Given:}$

$\mathsf{\dfrac{1}{1-x}}$

$\textbf{To find:}$

$\textsf{n th term in the expansion of}\;\mathsf{\dfrac{1}{1-x}}$

$\textbf{Solution:}$

$\textsf{Let}\;\mathsf{f(x)=\dfrac{1}{1-x}=(1-x)^{-1}\;\implies\;f(0)=1}$

$\mathsf{f^1(x)=(1-x)^{-2}\;\implies\;f^1(0)=1=1!}$

$\mathsf{f^2(x)=2(1-x)^{-3}\;\implies\;f^2(0)=2=2!}$

$\mathsf{f^3(x)=6(1-x)^{-4}\;\implies\;f^3(0)=6=3!}$

$\mathsf{f^4(x)=24(1-x)^{-5}\;\implies\;f^4(0)=24=4!}$

$\textsf{Proceeding like this, we get}$

$\mathsf{f^{n-1}(0)=(n-1)!}$

$\textsf{The maclaurin series of f(x) is}$

$\mathsf{f(x)=f(0)+\dfrac{f^1(0)}{1!}x^1+\dfrac{f^2(0)}{2!}x^2+\;.\;.\;.\;.\;.}$

$\mathsf{n\;th\;term,\;T_n=\dfrac{f^{n-1}(0)}{(n-1)!}}x^{n-1}}$

$\implies\mathsf{T_n=\dfrac{(n-1)!}{(n-1)!}}x^{n-1}}$

$\implies\boxed{\mathsf{T_n=x^{n-1}}}$

$\textbf{Find more:}$

First two terms in expansion of e^x.secx by maclaurins theorem is

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