Question

The coefficient of x5 in the Maclaurin's expansion of In(1 + x) is​

Answer 1

Answer:

(x+1)+

5+1

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Answer 2

A Maclaurin series is a Taylor series expansion of a function about 0.

Given function:

$f(x)=ln(1+x)$

Using $x=0,$ the given equation function becomes $f(0)=ln(1+0)=ln1=0$

$f'(x)=\frac{1}{1+x}; f'(0)=1\\ f''(x)=\frac{1}{(1+x)^{2} };f''(0)=-1\\ f'''(x)=\frac{2}{(1+x)^{3} };f'''(0)=2$

$f^{4}(x)=\frac{-6}{(1+x)^{4} };f^{4}(0)=-6\\ f^{5} (x)=\frac{24}{(1+x)^{5} } ;f^{5}(0)=24$

Maclaurin series at $x=0.$

$=0+\frac{1}{1!}x+\frac{-1}{2!}x^{2}+\frac{2}{3!}x^{3}+\frac{-6}{4!}x^{4}+\frac{24}{5!}x^{5}+...$

$=x-\frac{1}{2}x^{2} +\frac{1}{3}x^{3}-\frac{1}{4}x^{4}+\frac{1}{5}x^{5}$

Hence, the coefficient of $x^{5}$ is $\frac{1}{5}.$