Question

The mean of the numbers $\frac{{}^{50}{C}_0}{1},\frac{{}^{50}{C}_2}{3},\frac{{}^{50}{C}_4}{5},...,\frac{{}^{50}{C}_{50}}{51}$ equals :

• A. $\frac{2^{50}}{51}$
• B. $\frac{2^{49}}{51}$
• C. $\frac{2^{49}}{39\times 17}$
• D. None of these

Answer 1

Consider,

${(1+x)}^{50}{=}^{50}{C}_0{+}^{50}{C}_1{x}^1+...{+}^{50}{C}_{50}{x}^{50}$ ...(1)

${(1-x)}^{50}{=}^{50}{C}_0{-}^{50}{C}_1{x}^1+...{+}^{50}{C}_{50}{x}^{50}$ ...(2)

Adding (1) and (2),we get

$2{(}^{50}{C}_0{+}^{50}{C}_2{x}^2+...{+}^{50}{C}_{50}{x}^{50})={(1+x)}^{50}+{(1-x)}^{50}$

${(}^{50}{C}_0{+}^{50}{C}_2{x}^2+...{+}^{50}{C}_{50}{x}^{50})=\frac{1}{2}{(1+x)}^{50}+{(1-x)}^{50}$

Integrating with limits $0$ to $1,$ we get

${\int }_0^1{(}^{50}{C}_0{+}^{50}{C}_2{x}^2+...{+}^{50}{C}_{50}{x}^{50})={\int }_0^1(\frac{1}{2}{(1+x)}^{50}+{(1-x)}^{50})$

$\Rightarrow {[{}^{50}{C}_{0}x+\frac{{}^{50}{C}_2}{3}{x}^3+\frac{{}^{50}{C}_4}{5}{x}^5+...+\frac{{}^{50}{C}_{50}{x}^{51}}{51}]}_0^1$

$=\frac{1}{2}{[\frac{{(1+x)}^{51}}{51}-\frac{{(1-x)}^{51}}{51}]}_0^1$

$\Rightarrow \left[{}^{50}{C}_0\right(1)+\frac{{}^{50}{C}_2}{3}{\left(1\right)}^3+\frac{{}^{50}{C}_4}{5}{\left(1\right)}^5+...+\frac{{}^{50}{C}_{50}{\left(1\right)}^{51}}{51}]-\left[{}^{50}{C}_0\right(0)+\frac{{}^{50}{C}_2}{3}{\left(0\right)}^3+\frac{{}^{50}{C}_4}{5}{\left(0\right)}^5+...+\frac{{}^{50}{C}_{50}{\left(0\right)}^{51}}{51}]=\frac{1}{2}[\frac{{(1+1)}^{51}}{51}-\frac{{(1-1)}^{51}}{51}]-\frac{1}{2}[\frac{{(1+0)}^{51}}{51}-\frac{{(1-0)}^{51}}{51}]$

${\Rightarrow }^{50}{C}_0+\frac{{}^{50}{C}_2}{3}+\frac{{}^{50}{C}_4}{5}+...\frac{{}^{50}{C}_{50}}{51}=\frac{1}{2}\times \frac{2^{51}}{51}=\frac{2^{50}}{51}$

The number of even terms from $0$ to $50$ are $\frac{1}{2}(51+1)=26$

[Since, Number of terms from $0$ to $50$ are $51$--(odd)]

$\Rightarrow$ Number of terms from ${}^{50}{C}_0$ to ${}^{50}{C}_{50}$ are $26$ .

$\therefore$ Required mean

$\frac{\frac{{{}^{50}C}_0}{1}+\frac{{}^{50}{C}_2}{3}+\frac{{}^{50}{C}_4}{5}+...+\frac{{{}^{50}C}_{50}}{51}}{26}$

$=\frac{2^{50}}{51}\times \frac{1}{26}$

$=\frac{2^{49}\times 2}{3\times 17\times 26}$

$=\frac{2^{49}}{39\times 17}$

Hence, the correct option is C.