Question

If $C_0,C_1,C_2,\dots ,C_{15}$ are binomial coefficients in ${(1+x)}^{15}$, then $\frac{C_1}{C_0}+2\frac{C_2}{C_1}+3\frac{C_3}{C_2}+\cdots +15\frac{C_{15}}{C_{14}}$ is equal to

• A. $60$
• B. $120$
• C. $64$
• D. $124$
• E. $144$

Answer 1

The correct option is B $120$

We know that,
$\frac{{}^n{C}_r}{{}^n{C}_{r-1}}=\frac{n-(r-1)}{r}$

$\Rightarrow r\cdot \frac{{}^n{C}_r}{{}^n{C}_{r-1}}=n+1-r$

$\Rightarrow \sum _{r=1}^{n}r\cdot \frac{{}^{16}{C}_r}{{}^{16}{C}_{r-1}}=\sum _{r=1}^{16}(16-r)$

$16\times 15-\sum _{r=1}^{n}r$

$=16\times 15-\frac{15\times 16}{2}$

$=240-120$

$=120$