Let the sum of the coefficients of the first three terms in the expansion of be 376. Then the coefficient of x4 is ___________ . [2023]

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Solution

Q.
Let the sum of the coefficients of the first three terms in the expansion of ${(x - \frac{3}{x^{2}})}^{n}, x \neq 0, n \in N,$ be 376. Then the coefficient of $x^{4}$ is ___________ . [2023]

Ans.
(405)

Sum of coefficients of first three terms of ${(x - \frac{3}{x^{2}})}^{n}$ is 376.

${}^{n}C_{0} - {}^{n}C_{1} \cdot 3 + {}^{n}C_{2} \cdot 3^{2} = 376 \Rightarrow 1 - 3 n + \frac{n (n - 1)}{2} \times 9 = 376$

$\Rightarrow 3 n^{2} - 5 n - 250 = 0 ∴ n = 10$

Now, $T_{r + 1} = {}^{10}C_{r} x^{10 - r} {(\frac{- 3}{x^{2}})}^{r} = {}^{10}C_{r} {(- 3)}^{r} \cdot x^{10 - 3 r}$

For coefficient of $x^{4}$ $\Rightarrow 10 - 3 r = 4 \Rightarrow r = 2$

Coefficient of $x^{4}$ $= {}^{10}C_{2} {(- 3)}^{2} = 405$

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