Let the sum of the coefficients of the first three terms in the expansion of ${\left( {x - {3 \over {{x^2}}}} \right)^n},x \ne 0.~n \in \mathbb{N}$, be 376. Then the coefficient of $x^4$ is __________.
Answer (integer)
405
Solution
$S=1-3 n+\frac{9 n(n-1)}{2}=376$
<br/><br/>
$$
\begin{aligned}
& 3 n^{2}-5 n-250=0 \\\\
& n=10, \frac{-25}{3} \text { (Rejected) } \\\\
& T_{r+1}={ }^{n} C_{r} \cdot x^{n-r}\left(\frac{-3}{x^{2}}\right)^{r} \\\\
& ={ }^{n} C_{r} x x^{n-3 r}(-3)^{r} \\\\
& ={ }^{10} C_{r} x^{10-3 r}(-3)^{r}
\end{aligned}
$$
<br/><br/>
Here $r=2$
<br/><br/>
$$
\begin{aligned}
\text {Required coefficient } & ={ }^{10} \mathrm{C}_{2}(-3)^{2} \\\\
& =45 \times 9 \\\\
& =405
\end{aligned}
$$
About this question
Subject: Mathematics · Chapter: Binomial Theorem · Topic: Binomial Expansion
This question is part of PrepWiser's free JEE Main question bank. 193 more solved questions on Binomial Theorem are available — start with the harder ones if your accuracy is >70%.