Question

The mean value of the numbers $\frac{{}^{30}{C}_0}{1},\frac{{}^{30}{C}_2}{3},.....,\frac{{}^{30}{C}_{20}}{21},....,\frac{{}^{30}{C}_{22}}{23},....,\frac{{}^{30}{C}_{30}}{31}$ equals

• A. $\frac{2^{31}}{31}$
• B. $\frac{4^{13}}{31}$
• C. $\frac{2^{13}}{31}$
• D. None of these

Answer 1

The correct option is B $\frac{4^{13}}{31}$

Consider ${(1+x)}^{30}{=}^{30}{C}_0{+}^{30}{C}_1{x}^1{+}^{30}{C}_2{x}^2+...{+}^{30}{C}_{30}{x}^{30}...\left(i\right)$
and ${(1-x)}^{30}{=}^{30}{C}_0{-}^{30}{C}_1{x}^1{+}^{30}{C}_2{x}^2+...{+}^{30}{C}_{30}{x}^{30}...\left(ii\right)$
adding (i) and (ii) we get
${}^{30}{C}_0{+}^{30}{C}_2{x}^2{+}^{30}{C}_4{x}^4+...+...{+}^{30}{C}_{30}{x}^{30}$
$=\frac{1}{2}[{(1+x)}^{30}+{(1-x)}^{30}]$
integrating with respect to x with limit 0 to 1, we get
${}^{30}{C}_0+\frac{{}^{30}{C}_2}{3}+\frac{{}^{30}{C}_4}{5}+.....+\frac{{}^{30}{C}_{30}}{31}$$=\frac{1}{2}{[\frac{{(1+x)}^{31}}{31}-\frac{{(1-x)}^{31}}{31}]}_0^1$ $=\frac{1}{2}\frac{2^{31}}{31}$ ${}^{30}{C}_0+\frac{{}^{30}{C}_2}{3}+\frac{{}^{30}{C}_4}{5}.....+\frac{{}^{30}{C}_{30}}{31}=\frac{2^{30}}{31}$
Now number of numbers from $\frac{{}^{30}{C}_0}{1},\frac{{}^{30}{C}_2}{3}.....\frac{{}^{30}{C}_{30}}{31}$ are 16
$\therefore$ Mean of these numbers is
$=\frac{\frac{{}^{30}{C}_0}{1}+\frac{{}^{30}{C}_2}{3}+\frac{{}^{30}{C}_4}{5}+....+\frac{{}^{30}{C}_{30}}{31}}{16}$
$=\frac{2^{30}}{31\times 16}=\frac{2^{26}}{31}=\frac{4^{13}}{31}.$