Question

Find ${}^{(2n+1)}{C}_1{+}^{(2n+1)}{C}_3+\dots \dots {+}^{(2n+1)}{C}_{2n+1}=$

• A. $2^{2n+1}$
• B. $2^{2n+1}-1$
• C. $2^{2n}$
• D. $2^{2n}-1$

Answer 1

The correct option is C $2^{2n}$

Consider
$(1+x{)}^{2n+1}={}^{2n+1}{C}_0+{}^{2n+1}{C}_{1}x+{}^{2n+1}{C}_2{x}^2+...{}^{2n+1}{C}_{2n+1}{x}^{2n+1}$
Substituting, $x=1$, we get
$2^{2n+1}={}^{2n+1}{C}_0+{}^{2n+1}{C}_1+{}^{2n+1}{C}_2+...{}^{2n+1}{C}_{2n+1}$
Substituting $x=-1$ we get
$0={}^{2n+1}{C}_0-{}^{2n+1}{C}_1+{}^{2n+1}{C}_2+...-{}^{2n+1}{C}_{2n+1}$
Subtracting (ii) from (i), we get
$2^{2n+1}=2[{}^{2n+1}{C}_1+{}^{2n+1}{C}_3+{}^{2n+1}{C}_5+...{}^{2n+1}{C}_{2n+1}]$
$2^{2n}=[{}^{2n+1}{C}_1+{}^{2n+1}{C}_3+{}^{2n+1}{C}_5+...{}^{2n+1}{C}_{2n+1}]$