Question

Let $C _{ r }$ 's denote the combinational coefficients in the expansion of the binomial $(1+ x )^{14}$, then the value of the sum $\displaystyle\sum_{ r =1}^{15} r \cdot C _{ r -1}$ 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+\ldots \ldots .+15 \cdot C _{14}$
$Re$-writing the sum as reverse order
$S =15 \cdot C _0+14 C _1+13 C _2+\ldots \ldots+ C _{14^*}$
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$2 S =16\left[ C _0+ C _1+ C _2+\ldots \ldots+ C _{14}\right]=16 \cdot 2^{14} $
$S =2^3 \cdot 2^{14}=2^{17} \Rightarrow( B )$