Question

The value of the sum of the series ${}^{14}{C}_0\cdot {}^{15}{C}_1+{}^{14}{C}_1\cdot {}^{15}{C}_2$ $+{}^{14}{C}_2\cdot {}^{15}{C}_3+\dots +{}^{14}{C}_{14}{}^{15}{C}_{15}$ is

• A. ${}^{29}{C}_{12}$
• B. ${}^{29}{C}_{10}$
• C. ${}^{29}{C}_{14}$
• D. ${}^{29}{C}_{16}$

Answer 1

Answer: (C)

$(1+x{)}^{14}={}^{14}{C}_0+{}^{14}{C}_{1}x+{}^{14}{C}_2{x}^2+\dots +{}^{14}{C}_{14}{x}^{14}$ ...(1)
and $(x+1){}^{15}={}^{15}{C}_0{x}^{15}+{}^{15}{C}_1{x}^{14}+{}^{15}{C}_2{x}^{13}+{}^{15}{C}_3{x}^{12}+\dots +{}^{15}{C}_{15}$ ...(2)
Multiplying (1) and (2) and equating the coefficients of $x^{14}$, we get
${}^{14}{C}_0\cdot {}^{15}{C}_1+{}^{14}{C}_1\cdot {}^{15}{C}_2+{}^{14}{C}_2\cdot {}^{15}{C}_3+\dots +{}^{14}{C}_{14}\cdot {}^{15}{C}_{15}$
=the coefficient of $x^{14}$ in $(1+x{)}^{29}={}^{29}{C}_{14}$