Question

Let $(1+x{)}^{10}=\sum _{r=0}^{10}{c}_r{x}^r$ and $(1+x{)}^7=\sum _{r=0}^7{d}_r{x}^r$. If $P=\sum _{r=0}^5{c}_{2r}$ and $Q=\sum _{r=0}^3{d}_{2r+1}$, then $\frac{P}{Q}$ is equal to

• A. $4$
• B. $8$
• C. $16$
• D. $32$

Answer 1

Answer: (B)

$8$

$P=\sum _{r=0}^5{C}_{2r}$
$={}^{10}{C}_0+{}^{10}{C}_2+\cdots +{}^{10}{C}_{10}=\frac{2^{10}}{2}=2^9$

$Q=\sum _{r=0}^3{d}_{2r+1}=d_1+d_3+d_5+d_7$
$={}^7{C}_1+{}^7{C}_3+{}^7{C}_5+{}^7{C}_7=\frac{2^7}{2}=2^6$
$\therefore \frac{P}{Q}=\frac{2^9}{2^6}=2^3=8$