The natural number m,for which the coefficien | vedclass

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(C) The general term in the expansion of $\left( x^{m} + x^{-2} \right)^{22}$ is given by $T_{r+1} = {}^{22}C_{r} (x^{m})^{22-r} (x^{-2})^{r} = {}^{22

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$13$

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(C) The general term in the expansion of $\left( x^{m} + x^{-2} ight)^{22}$ is given by $T_{r+1} = {}^{22}C_{r} (x^{m})^{22-r} (x^{-2})^{r} = {}^{22}C_{r} x^{22m - mr - 2r}$.For the coefficient of $x$,we set the exponent of $x$ equal to $1$:$22m - mr - 2r = 1 \implies r(m+2) = 22m - 1$.We are given that the coefficient is $1540$,so ${}^{22}C_{r} = 1540$.Since ${}^{22}C_{3} = \frac{22 	imes 21 	imes 20}{3 	imes 2 	imes 1} = 1540$,we have $r = 3$ or $r = 22 - 3 = 19$.Case $1$: If $r = 3$,then $3(m+2) = 22m - 1 \implies 3m + 6 = 22m - 1 \implies 19m = 7$,which gives $m = \frac{7}{19}$,not a natural number.Case $2$: If $r = 19$,then $19(m+2) = 22m - 1 \implies 19m + 38 = 22m - 1 \implies 3m = 39 \implies m = 13$.Thus,the natural number $m$ is $13$.

The natural number $m$,for which the coefficient of $x$ in the binomial expansion of $\left( x^{m} + \frac{1}{x^{2}} \right)^{22}$ is $1540$,is

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