For a positive integer n,(1+frac{1}{x})^n is | vedclass

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(D) The expansion of $(1+\frac{1}{x})^n$ is $\sum_{k=0}^{n} {}^{n}C_k (\frac{1}{x})^k = \sum_{k=0}^{n} {}^{n}C_k x^{-k}$.Since the expansion is in inc

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$118$

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(D) The expansion of $(1+\frac{1}{x})^n$ is $\sum_{k=0}^{n} {}^{n}C_k (\frac{1}{x})^k = \sum_{k=0}^{n} {}^{n}C_k x^{-k}$.Since the expansion is in increasing powers of $x$,we consider the coefficients of $x^{-r-1}, x^{-r}, x^{-r+1}$ which are ${}^{n}C_{r+1}, {}^{n}C_r, {}^{n}C_{r-1}$.Given the ratio ${}^{n}C_{r-1} : {}^{n}C_r : {}^{n}C_{r+1} = 2 : 5 : 12$.From $\frac{{}^{n}C_{r-1}}{{}^{n}C_r} = \frac{2}{5}$,we get $\frac{r}{n-r+1} = \frac{2}{5}$ $\Rightarrow 5r = 2n - 2r + 2$ $\Rightarrow 7r = 2n + 2$ (Equation $1$).From $\frac{{}^{n}C_r}{{}^{n}C_{r+1}} = \frac{5}{12}$,we get $\frac{r+1}{n-r} = \frac{5}{12}$ $\Rightarrow 12r + 12 = 5n - 5r$ $\Rightarrow 17r = 5n - 12$ (Equation $2$).Multiplying Equation $1$ by $17$ and Equation $2$ by $7$:$119r = 34n + 34$$119r = 35n - 84$Equating the two: $35n - 84 = 34n + 34 \Rightarrow n = 118$.

For a positive integer $n$,$(1+\frac{1}{x})^n$ is expanded in increasing powers of $x$. If three consecutive coefficients in this expansion are in the ratio $2:5:12$,then $n$ is equal to

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