If the coefficients of x^7 and x^8 in the exp | vedclass

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(C) The general term in the expansion of $[2 + \frac{x}{3}]^n$ is given by $T_{r+1} = ^nC_r (2)^{n-r} (\frac{x}{3})^r = ^nC_r (2)^{n-r} (\frac{1}{3})^

Vedclass · Accepted Answer

$55$

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(C) The general term in the expansion of $[2 + \frac{x}{3}]^n$ is given by $T_{r+1} = ^nC_r (2)^{n-r} (\frac{x}{3})^r = ^nC_r (2)^{n-r} (\frac{1}{3})^r x^r$.The coefficient of $x^7$ is $^nC_7 (2)^{n-7} (\frac{1}{3})^7$.The coefficient of $x^8$ is $^nC_8 (2)^{n-8} (\frac{1}{3})^8$.Given that these coefficients are equal:$^nC_7 (2)^{n-7} (\frac{1}{3})^7 = ^nC_8 (2)^{n-8} (\frac{1}{3})^8$.Dividing both sides by $^nC_7 (2)^{n-8} (\frac{1}{3})^7$,we get:$2 = \frac{^nC_8}{^nC_7} 	imes \frac{1}{3}$.Using the property $\frac{^nC_r}{^nC_{r-1}} = \frac{n-r+1}{r}$,we have $\frac{^nC_8}{^nC_7} = \frac{n-8+1}{8} = \frac{n-7}{8}$.So,$2 = \frac{n-7}{8} 	imes \frac{1}{3} \Rightarrow 2 = \frac{n-7}{24}$.$n - 7 = 48 \Rightarrow n = 55$.

If the coefficients of $x^7$ and $x^8$ in the expansion of $[2 + \frac{x}{3}]^n$ are equal,then the value of $n$ is:

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