Question

Let $C_r$ 's denote the combinational coefficients in the expansion of the binomial $(1+x{)}^{14}$, then the value of the sum $\sum _{r=1}^{15}r\cdot C_{r-1}$ is

• A. $2^{15}$
• B. $2^{16}$
• C. $2^{17}$
• D. $2^{18}$

Answer 1

Answer: (C)

$S=1\cdot C_0+2\cdot C_1+3\cdot C_2+\dots \dots .+15\cdot C_{14}$
$Re$-writing the sum as reverse order
$S=15\cdot C_0+14C_1+13C_2+\dots \dots +C_{14^{\ast }}$
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$2S=16\left[C_0+C_1+C_2+\dots \dots +C_{14}\right]=16\cdot 2^{14}$
$S=2^3\cdot 2^{14}=2^{17}\Rightarrow (B)$