Find a if the 17^{text{th}} and 18^{text{th}} | vedclass

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(A) The $(r+1)^{	ext{th}}$ term of the expansion $(x+y)^n$ is given by $T_{r+1} = {^nC_r} x^{n-r} y^r$.For the $17^{	ext{th}}$ term,we have $r+1 = 1

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(A) The $(r+1)^{	ext{th}}$ term of the expansion $(x+y)^n$ is given by $T_{r+1} = {^nC_r} x^{n-r} y^r$.For the $17^{	ext{th}}$ term,we have $r+1 = 17$,i.e.,$r = 16$.Therefore,$T_{17} = T_{16+1} = {^{50}C_{16}} (2)^{50-16} a^{16} = {^{50}C_{16}} 2^{34} a^{16}$.Similarly,for the $18^{	ext{th}}$ term,$r = 17$.Therefore,$T_{18} = T_{17+1} = {^{50}C_{17}} (2)^{50-17} a^{17} = {^{50}C_{17}} 2^{33} a^{17}$.Given that $T_{17} = T_{18}$,we have:${^{50}C_{16}} 2^{34} a^{16} = {^{50}C_{17}} 2^{33} a^{17}$.Dividing both sides by ${^{50}C_{17}} 2^{33} a^{16}$,we get:$a = \frac{{^{50}C_{16}} \cdot 2^{34}}{{^{50}C_{17}} \cdot 2^{33}} = \frac{{^{50}C_{16}}}{{^{50}C_{17}}} \cdot 2$.Using the formula $\frac{{^nC_r}}{{^nC_{r-1}}} = \frac{n-r+1}{r}$,where $n=50$ and $r=17$:$\frac{{^{50}C_{16}}}{{^{50}C_{17}}} = \frac{17}{50-17+1} = \frac{17}{34} = \frac{1}{2}$.Thus,$a = \frac{1}{2} 	imes 2 = 1$.

Find $a$ if the $17^{\text{th}}$ and $18^{\text{th}}$ terms of the expansion $(2 + a)^{50}$ are equal.

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