If the 17^{text{th}} and the 18^{text{th}} te | vedclass

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(D) Given that the $17^{	ext{th}}$ and $18^{	ext{th}}$ terms in the expansion of $(2+a)^{50}$ are equal.The general term $T_{r+1}$ in the expansion

Vedclass · Accepted Answer

-$36$

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(D) Given that the $17^{	ext{th}}$ and $18^{	ext{th}}$ terms in the expansion of $(2+a)^{50}$ are equal.The general term $T_{r+1}$ in the expansion of $(x+y)^n$ is given by $^{n}C_{r} x^{n-r} y^{r}$.For $T_{17} = T_{18}$:$^{50}C_{16} (2)^{50-16} (a)^{16} = ^{50}C_{17} (2)^{50-17} (a)^{17}$$^{50}C_{16} (2)^{34} (a)^{16} = ^{50}C_{17} (2)^{33} (a)^{17}$Dividing both sides by $^{50}C_{16} (2)^{33} (a)^{16}$:$2 = \frac{^{50}C_{17}}{^{50}C_{16}} 	imes a$Using the property $\frac{^{n}C_{r}}{^{n}C_{r-1}} = \frac{n-r+1}{r}$:$2 = \frac{50-17+1}{17} 	imes a = \frac{34}{17} 	imes a = 2a$Thus,$a = 1$.Now,we need the coefficient of $x^{35}$ in the expansion of $(a+x)^{-2} = (1+x)^{-2}$.The expansion of $(1+x)^{-n} = 1 - nx + \frac{n(n+1)}{2!} x^2 - \dots + (-1)^{r} \frac{(n+r-1)!}{(n-1)! r!} x^r + \dots$For $(1+x)^{-2}$,the coefficient of $x^r$ is $(-1)^r \frac{(2+r-1)!}{(2-1)! r!} = (-1)^r (r+1)$.For $r=35$,the coefficient is $(-1)^{35} (35+1) = -36$.

If the $17^{\text{th}}$ and the $18^{\text{th}}$ terms in the expansion of $(2+a)^{50}$ are equal,then the coefficient of $x^{35}$ in the expansion of $(a+x)^{-2}$ is

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