Question

Find 'a' if the 17th and 18th term in the expansion of ${(2+a)}^{50}$ as equal.

Answer 1

5)We know that General term of expansion ${(a+b)}^n$ is

$T_{r+1}{=}^n{C}_r{a}^{n-r}{b}^r$ .......$\left(1\right)$

$T_{17}=T_{16+1}$ of ${(2+a)}^{50}$

Put $r=16,n=50,a=2$ and $b=a$ in $\left(1\right)$

$T_{16+1}{=}^{50}{C}_{16}{\left(2\right)}^{50-16}{a}^{16}{=}^{50}{C}_{16}{\left(2\right)}^{34}{a}^{16}$

$T_{18}=T_{17+1}$ of ${(2+a)}^{50}$

Put $r=17,n=50,a=2$ and $b=a$ in $\left(1\right)$

$T_{17+1}{=}^{50}{C}_{17}{\left(2\right)}^{50-17}{a}^{17}{=}^{50}{C}_{17}{\left(2\right)}^{33}{a}^{17}$

It is given that $T_{17}=T_{18}$

${\Rightarrow }^{50}{C}_{16}{\left(2\right)}^{34}{a}^{16}{=}^{50}{C}_{17}{\left(2\right)}^{33}{a}^{17}$

$\Rightarrow a=\frac{{}^{50}{C}_{16}}{{}^{50}{C}_{17}}{\left(2\right)}^{34-33}$

$\Rightarrow a=\frac{50!}{16!\times 34!}\times \frac{17!\times 33!}{50!}\times 2$ using ${}^n{C}_r=\frac{n!}{r!(n-r)!}$

$\Rightarrow a=\frac{17}{34}\times 2=1$

Hence $a=1$