Question

If $(1+x{)}^{15}=a_0+a_{1}x+\dots +a_{15}{x}^{15},$ then $\sum _{r=1}^{15}r\frac{a_r}{a_r-1}$ is equal to

• A. 110
• B. 115
• C. 120
• D. 135

Answer 1

Answer: (C)

Given that
$(1+x{)}^{15}=a_0+a_{1}x+a_2{x}^2+\dots +a_{15}{x}^{15}$
$\Rightarrow {}^{15}{C}_0+{}^{15}{C}_{1}x+{}^{15}{C}_2{x}^2+\dots +{}^{15}{C}_{15}{x}^{15}$
$=a_0+a_{1}x+a_2{x}^2+\dots +a_{15}{x}^{15}$
Equating the coefficient of various powers of $x$, we get
$a_0={}^{15}{C}_0,a_1={}^{15}{C}_1,a_2={}^{15}{C}_2,\dots ,a_{15}={}^{15}{C}_{15}$
$\therefore \sum _{r=1}^{15}r\frac{a_r}{a_{r-1}}=\sum _{r=1}^{15}r\frac{{}^{15}{C}_r}{{}^{15}{C}_{r-1}}$
$=\sum _{r=1}^{15}r\frac{\frac{15!}{r!(15-r)!}}{\frac{15!}{(r-1)!(15-r+1)!}}$
$=\sum _{r=1}^{15}\frac{(r-1)!(15-r+1)!}{r!(15-r)!}$
$=\sum _{r=1}^{15}15-r+1=15+14+13+\dots +2+1$
$=\frac{15(15+1)}{2}=120$