Search any question & find its solution
Question:
Answered & Verified by Expert
Coefficient of $x^{10}$ in the expansion of $(2+3 x) e^{-x}$ is
Options:
Solution:
2133 Upvotes
Verified Answer
The correct answer is:
$\frac{-28}{(10) !}$
$$
\begin{aligned}
(2+3 x) e^{-x} & \\
& =(2+3 x)\left(1-\frac{x}{1 !}+\frac{x^2}{2 !}-\frac{x^3}{3 !}+\ldots\right)
\end{aligned}
$$
$\therefore$ Coefficient of $x^{10}$ in the above series
$$
\begin{aligned}
& =\frac{2}{10 !}-\frac{3}{9 !} \\
& =\frac{1}{10 !}(2-30)=\frac{-28}{10 !}
\end{aligned}
$$
\begin{aligned}
(2+3 x) e^{-x} & \\
& =(2+3 x)\left(1-\frac{x}{1 !}+\frac{x^2}{2 !}-\frac{x^3}{3 !}+\ldots\right)
\end{aligned}
$$
$\therefore$ Coefficient of $x^{10}$ in the above series
$$
\begin{aligned}
& =\frac{2}{10 !}-\frac{3}{9 !} \\
& =\frac{1}{10 !}(2-30)=\frac{-28}{10 !}
\end{aligned}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.