Question

Mean of numbers $\frac{{}^{50}{C}_0}{1},\frac{{}^{50}{C}_2}{3},\frac{{}^{50}{C}_4}{5},\cdots ,\frac{{}^{50}{C}_{50}}{51}$ is

• A. $\frac{2^{49}}{663}$
• B. $\frac{2^{50}}{663}$
• C. $\frac{2^{49}}{51}$
• D. $\frac{2^{49}}{54}$

Answer 1

The correct option is A $\frac{2^{49}}{663}$

From the expansion of binomial
$(1+x{)}^{50}=C_0+C_{1}x+C_2{x}^2+C_3{x}^3+\cdots +C_{50}{x}^{50}$
integrating both sides w.r.t $x$ from the limits $-1$ to $1$, we get
${\left[\frac{(1+x{)}^{51}}{51}\right]}_{-1}^1={[C_{0}x+\frac{C_1{x}^2}{2}+\frac{C_2{x}^3}{3}+\cdots +\frac{C_{50}{x}^{51}}{51}]}_{-1}^1$
Now applying the limits, we get
$\frac{2^{51}}{51}=2[C_0+\frac{C_2}{3}+\frac{C_4}{5}+\cdots +\frac{C_{50}}{51}]$
$\therefore \text{Mean}=\frac{\text{sum of numbers}}{\text{total numbers}}=\frac{2^{50}}{51\times 26}=\frac{2^{49}}{663}$