Let be the constant term in the binomial expansion of . If the sum of the coefficients of the remaining terms in the expansion is 649 and the coefficient of is , then is equal to _________ . [2023]

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Solution

Q.
Let $α$ be the constant term in the binomial expansion of ${(\sqrt{x} - \frac{6}{x^{\frac{3}{2}}})}^{n}, n \leq 15$ . If the sum of the coefficients of the remaining terms in the expansion is 649 and the coefficient of $x^{- n}$ is $λ α$ , then $λ$ is equal to _________ . [2023]

Ans.
(36)

General term in the expansion of ${(\sqrt{x} - \frac{6}{x^{3 / 2}})}^{n}$ is given by
$T_{r + 1} = {}^{n}{C_{r}} {(x)}^{\frac{n - r}{2}} {(- 6)}^{r} {(x)}^{\frac{- 3}{2} r}$

For the constant term, the power of $x$ should be zero.

$\Rightarrow \frac{n - r}{2} - \frac{3}{2} r = 0 \Rightarrow n - 4 r = 0 \Rightarrow r = \frac{n}{4}$

For the constant term, put $x = 1$ in ${(\sqrt{x} - \frac{6}{x^{3 / 2}})}^{n}$

Constant term, ${(- 5)}^{n}$

$∴$ $Sum of coefficients except constant term is {(- 5)}^{n} - {}^{n}{C_{n / 4}} {(- 6)}^{n / 4} = 649$

Put $n = 4,$ $625 + 24 = 649$

Thus, $n = 4$ satisfies the above equation.

Now, $r = \frac{n}{4} = \frac{4}{4} = 1$

Required coefficient of $x^{- 4}$ = ${}^{4}{C_{3}} {(- 6)}^{3} = 4 (- 216)$

and constant term $a = {}^{4}{C_{1}} {(- 6)}^{1} = 4 (- 6)$

$∴$ $4 (- 216) = λ [4 (- 6)] \Rightarrow λ = 36$

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