Question

If ${}^{2n+1}{C}_0+{}^{2n+1}{C}_1+{}^{2n+1}{C}_2+\cdots +{}^{2n+1}{C}_n=4^{11}$ then which of the following is/are correct ?

• A. ${}^n{C}_0+\frac{{}^n{C}_1}{2}+\frac{{}^n{C}_2}{2^2}+\cdots \frac{{}^n{C}_n}{2^n}={\left(\frac{3}{2}\right)}^{11}$
• B. ${}^n{C}_0+\frac{{}^n{C}_1}{2}+\frac{{}^n{C}_2}{2^2}+\cdots \frac{{}^n{C}_n}{2^n}={\left(\frac{2}{3}\right)}^{11}$
• C. ${}^n{C}_0-\frac{{}^n{C}_1}{2}+\frac{{}^n{C}_2}{2^2}-\cdots +(-1{)}^n\frac{{}^n{C}_n}{2^n}={\left(\frac{1}{2}\right)}^{11}$
• D. ${}^n{C}_0-\frac{{}^n{C}_1}{2}+\frac{{}^n{C}_2}{2^2}-\cdots +(-1{)}^n\frac{{}^n{C}_n}{2^n}={\left(\frac{2}{3}\right)}^{11}$

Answer 1

Answer: (A), (C)

${}^n{C}_0-\frac{{}^n{C}_1}{2}+\frac{{}^n{C}_2}{2^2}-\cdots +(-1{)}^n\frac{{}^n{C}_n}{2^n}={\left(\frac{1}{2}\right)}^{11}$

${}^{2n+1}{C}_0+{}^{2n+1}{C}_1+{}^{2n+1}{C}_2+\cdots +{}^{2n+1}{C}_n=4^{11}$
$\Rightarrow \frac{1}{2}\left(2^{2n+1}\right)=2^{22}$
$\Rightarrow n=11$

Now,
${}^n{C}_0+\frac{{}^n{C}_1}{2}+\frac{{}^n{C}_2}{2^2}+\cdots +\frac{{}^n{C}_n}{2^n}=\sum _{r=0}^n\frac{{}^n{C}_r}{2^r}$
$=\sum _{r=0}^n{}^n{C}_r{\left(\frac{1}{2}\right)}^r={(1+\frac{1}{2})}^n={\left(\frac{3}{2}\right)}^n$

${}^n{C}_0-\frac{{}^n{C}_1}{2}+\frac{{}^n{C}_2}{2^2}+\cdots +(-1{)}^n\frac{{}^n{C}_n}{2^n}=\sum _{r=0}^n(-1{)}^r\frac{{}^n{C}_r}{2^r}$
$=\sum _{r=0}^n{}^n{C}_r{\left(\frac{-1}{2}\right)}^r={(1-\frac{1}{2})}^n={\left(\frac{1}{2}\right)}^n$