Question

The value of the sum of the series ${ }^{14} C_{0} \cdot{ }^{15} C_{1}+{ }^{14} C_{1} \cdot{ }^{15} C_{2}$ $+{ }^{14} C_{2} \cdot{ }^{15} C_{3}+\ldots+{ }^{14} C_{14}{ }^{15} C_{15}$ is

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

Answer 1

Answer: (C)

$(1+x)^{14}={ }^{14} C_{0}+{ }^{14} C_{1} x+{ }^{14} C_{2} x^{2}+\ldots+{ }^{14} C_{14} x^{14}$ ...(1)
and $(x+1){ }^{15}={ }^{15} C_{0} x^{15}+{ }^{15} C_{1} x^{14}+{ }^{15} C_{2} x^{13}+{ }^{15} C_{3} x^{12}+\ldots +{ }^{15} C_{15}$ ...(2)
Multiplying (1) and (2) and equating the coefficients of $x^{14}$, we get
${ }^{14} C_{0} \cdot{ }^{15} C_{1}+{ }^{14} C_{1} \cdot{ }^{15} C_{2}+{ }^{14} C_{2} \cdot{ }^{15} C_{3}+\ldots+{ }^{14} C_{14} \cdot{ }^{15} C_{15}$
=the coefficient of $x^{14}$ in $(1+x)^{29}={ }^{29} C_{14}$