Question

If $\sum _{k=0}^{18}\frac{k}{{{}^{18}C}_k}=a\sum _{k=0}^{18}\frac{1}{{{}^{18}C}_k}$, then the value of $a$ is

• A. $3$
• B. $9$
• C. $6$
• D. $18$
• E. $36$

Answer 1

The correct option is B $9$

Given $\sum _{k=0}^{18}\frac{k}{{{}^{18}C}_k}=a\sum _{k=0}^{18}\frac{1}{{{}^{18}C}_k}$
$LHS=\sum _{k=0}^{18}\frac{k}{{{}^{18}C}_k}=[0+\frac{1}{{{}^{18}C}_1}+\frac{1}{{{}^{18}C}_2}+...+\frac{1}{{{}^{18}C}_{18}}]...\left(i\right)$
$RHS=a\sum _{k=0}^{18}\frac{k}{{{}^{18}C}_k}=a[1+\frac{1}{{{}^{18}C}_1}+\frac{1}{{{}^{18}C}_2}+...+\frac{1}{{{}^{18}C}_{18}}]...\left(ii\right)$
From Eq(i) and (ii)
$18(1+\frac{1}{{{}^{18}C}_1}+\frac{1}{{{}^{18}C}_2}+...+\frac{1}{{{}^{18}C}_{18}})+\frac{9}{{{}^{18}C}_9}$
$=2a(1+\frac{1}{{{}^{18}C}_1}+\frac{1}{{{}^{18}C}_2}+...+\frac{1}{{{}^{18}C}_{18}})+\frac{a}{{{}^{18}C}_9}$
$\therefore$ $2a=18$
$\Rightarrow$ $a=9$