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

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

Vedclass · Accepted Answer

$55$

Vedclass · Answer

(B) The general term in the expansion of $(2+\frac{x}{3})^{n}$ is given by $T_{r+1} = {}^{n}C_{r} (2)^{n-r} (\frac{x}{3})^{r} = {}^{n}C_{r} (2)^{n-r} (\frac{1}{3})^{r} x^{r}$.The coefficient of $x^{7}$ is ${}^{n}C_{7} (2)^{n-7} (\frac{1}{3})^{7}$.The coefficient of $x^{8}$ is ${}^{n}C_{8} (2)^{n-8} (\frac{1}{3})^{8}$.Given that these coefficients are equal:${}^{n}C_{7} (2)^{n-7} (\frac{1}{3})^{7} = {}^{n}C_{8} (2)^{n-8} (\frac{1}{3})^{8}$.Dividing both sides by ${}^{n}C_{7} (2)^{n-8} (\frac{1}{3})^{7}$,we get:$2 = \frac{{}^{n}C_{8}}{{}^{n}C_{7}} \cdot \frac{1}{3}$.Using the property $\frac{{}^{n}C_{r}}{{}^{n}C_{r-1}} = \frac{n-r+1}{r}$,we have $\frac{{}^{n}C_{8}}{{}^{n}C_{7}} = \frac{n-8+1}{8} = \frac{n-7}{8}$.Substituting this into the equation:$2 = \frac{n-7}{8} \cdot \frac{1}{3} = \frac{n-7}{24}$.$n-7 = 48 \Rightarrow n = 55$.

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 equal to $.....$

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