Question

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

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

Answer 1

The correct option is C $\frac{2^{49}}{39\times 17}$

Consider

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

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

Adding (i) and (ii) we get,
${}^{50}{\text{C}}_0{+}^{50}{\text{C}}_2{x}^2{+}^{50}{\text{C}}_4{x}^4+...{+}^{50}{\text{C}}_{50}{x}^{50}=\frac{1}{2}[{(1+x)}^{50}+{(1-x)}^{50}]$

On integrating with limit $0$ to $1$ we get,
${[{}^{50}{\text{C}}_{0}x+\frac{{}^{50}{\text{C}}_2}{3}{x}^3+\frac{{}^{50}{\text{C}}_4}{5}{x}^5+...+\frac{{}^{50}{\text{C}}_{50}{x}^{51}}{51}]}_0^1=\frac{1}{2}{[\frac{{(1+x)}^{51}}{51}-\frac{{(1-x)}^{51}}{51}]}_0^1$
$\therefore$ ${}^{50}{\text{C}}_0+\frac{{}^{50}{\text{C}}_2}{3}+\frac{{}^{50}{\text{C}}_4}{5}+...+\frac{{}^{50}{\text{C}}_{50}}{51}=\frac{1}{2}\frac{2^{51}}{51}=\frac{2^{50}}{51}$

Now number of terms from ${}^{50}{\text{C}}_0$ to ${}^{50}{\text{C}}_{50}$ are $26$ (items)
$\therefore$ Required mean
$=\frac{{}^{50}{\text{C}}_0+\frac{{}^{50}{\text{C}}_2}{3}+\frac{{}^{50}{\text{C}}_4}{5}+...+\frac{{}^{50}{\text{C}}_{50}}{51}}{26}=\frac{2^{50}}{51}\times \frac{1}{26}$

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

Hence, option C.