JEE MAIN - Mathematics (2023 - 24th January Evening Shift - No. 18)
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
405
Explanation
$S=1-3 n+\frac{9 n(n-1)}{2}=376$
$$ \begin{aligned} & 3 n^{2}-5 n-250=0 \\\\ & n=10, \frac{-25}{3} \text { (Rejected) } \\\\ & T_{r+1}={ }^{n} C_{r} \cdot x^{n-r}\left(\frac{-3}{x^{2}}\right)^{r} \\\\ & ={ }^{n} C_{r} x x^{n-3 r}(-3)^{r} \\\\ & ={ }^{10} C_{r} x^{10-3 r}(-3)^{r} \end{aligned} $$
Here $r=2$
$$ \begin{aligned} \text {Required coefficient } & ={ }^{10} \mathrm{C}_{2}(-3)^{2} \\\\ & =45 \times 9 \\\\ & =405 \end{aligned} $$
$$ \begin{aligned} & 3 n^{2}-5 n-250=0 \\\\ & n=10, \frac{-25}{3} \text { (Rejected) } \\\\ & T_{r+1}={ }^{n} C_{r} \cdot x^{n-r}\left(\frac{-3}{x^{2}}\right)^{r} \\\\ & ={ }^{n} C_{r} x x^{n-3 r}(-3)^{r} \\\\ & ={ }^{10} C_{r} x^{10-3 r}(-3)^{r} \end{aligned} $$
Here $r=2$
$$ \begin{aligned} \text {Required coefficient } & ={ }^{10} \mathrm{C}_{2}(-3)^{2} \\\\ & =45 \times 9 \\\\ & =405 \end{aligned} $$
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