Question

The mean value of numbers ${}^{50}{C}_0{,}^{50}{C}_2{,}^{50}{C}_4,{\cdots }^{50}{C}_{50}$ is

• A. $\frac{2^{50}}{25}$
• B. $\frac{2^{48}}{13}$
• C. $\frac{2^{49}}{25}$
• D. None of these

Answer 1

The correct option is B $\frac{2^{48}}{13}$

Consider $(1+x{)}^{50}{=}^{50}{C}_0{+}^{50}{C}_{1}x{+}^{50}{C}_2{x}^2+.........{+}^{50}{C}_{50}{x}^{50}..(1)$
and $(1-x{)}^{50}{=}^{50}{C}_0{-}^{50}{C}_{1}x{+}^{50}{C}_2{x}^2+.........{+}^{50}{C}_{50}{x}^{50}..(2)$
Adding (1) and (2),
$(1+x{)}^{50}+(1-x{)}^{50}=2{(}^{50}{C}_0{+}^{50}{C}_2{x}^2{+}^{50}{C}_4{x}^4.........{+}^{50}{C}_{50}{x}^{50})..(3)$
Now put $x=1$ in (3), we get
${}^{50}{C}_0{+}^{50}{C}_2{+}^{50}{C}_4.........{+}^{50}{C}_{50}=\frac{2^{50}}{2}=2^{49}$

Since number of terms here is $26.$
Hence, the required mean $=\frac{{}^{50}{C}_0{+}^{50}{C}_2{+}^{50}{C}_4.........{+}^{50}{C}_{50}}{26}=\frac{2^{48}}{13}$