If the coefficients of x^7 and x^8 in (2 + fr | vedclass

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(B) 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}{

Vedclass · Accepted Answer

$55$

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(B) The general term in the expansion of $(2 + \frac{x}{3})^n$ is given by $T_{r+1} = {}^nC_r (2)^{n-r} (\frac{x}{3})^r = {}^nC_r (2)^{n-r} (\frac{1}{3})^r x^r$.For the coefficient of $x^7$,we set $r = 7$:Coefficient of $x^7 = {}^nC_7 (2)^{n-7} (\frac{1}{3})^7$.For the coefficient of $x^8$,we set $r = 8$:Coefficient of $x^8 = {}^nC_8 (2)^{n-8} (\frac{1}{3})^8$.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 ${}^nC_7 (2)^{n-8} (\frac{1}{3})^7$:$2 = \frac{{}^nC_8}{{}^nC_7} 	imes \frac{1}{3}$.Using the property $\frac{{}^nC_r}{{}^nC_{r-1}} = \frac{n-r+1}{r}$:$2 = \frac{n-8+1}{8} 	imes \frac{1}{3} = \frac{n-7}{24}$.$n - 7 = 48 \implies n = 55$.

If the coefficients of $x^7$ and $x^8$ in $(2 + \frac{x}{3})^n$ are equal,then $n$ is

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