Question

For Maclaurin series of $log(1+x)$, the coefficient of the third term is given by:

• A. $\frac{1}{3}$
• B. $-\frac{1}{3}$
• C. $\frac{2}{3}$
• D. $\frac{-2}{3}$

Answer 1

The correct option is A $\frac{1}{3}$

The Maclaurin series is given by $f\left(x\right)={\sum }_{k=0}^{\infty }\frac{f^{\left(k\right)}\left(a\right)}{k!}{x}^k$ where $a=0$
We have $f\left(x\right)=log(1+x)$
Since we have to find the coefficient of the third term, let us take $n=5$.
$\therefore f\left(x\right)\approx {\sum }_{k=0}^5\frac{f^{\left(k\right)}\left(0\right)}{k!}{x}^k$
$f^{\left(0\right)}\left(x\right)=log(1+x),\Rightarrow f^{\left(0\right)}\left(0\right)=0$
$f^{\left(1\right)}\left(x\right)=\frac{1}{x+1},\Rightarrow f^{\left(1\right)}\left(0\right)=1$
$f^{\left(2\right)}\left(x\right)=-\frac{1}{(x+1{)}^2},\Rightarrow f^{\left(2\right)}\left(0\right)=-1$
$f^{\left(3\right)}\left(x\right)=\frac{2}{(x+1{)}^3},\Rightarrow f^{\left(3\right)}\left(0\right)=2$
$f^{\left(4\right)}\left(x\right)=-\frac{6}{(x+1{)}^4},\Rightarrow f^{\left(4\right)}\left(0\right)=-6$
$f^{\left(5\right)}\left(x\right)=\frac{24}{(x+1{)}^5},\Rightarrow f^{\left(5\right)}\left(0\right)=24$
$\therefore f\left(x\right)\approx \frac{0}{0!}{x}^0+\frac{1}{1!}{x}^1+\frac{-1}{2!}{x}^2+\frac{2}{3!}{x}^3+\frac{-6}{4!}{x}^4+\frac{24}{5!}{x}^5$
$\Rightarrow f\left(x\right)\approx x-\frac{1}{2}{x}^2+\frac{1}{3}{x}^3-\frac{1}{4}{x}^4+\frac{1}{5}{x}^5$
Thus the coefficient of third term is $\frac{1}{3}$.