Question

If $C_{0},C_{1},C_{2},.....,C_{20}$ are the binomial coefficients in the expansion of $\left(1 + x\right)^{20},$ then the value of $\frac{C_{1}}{C_{0}}+2\frac{C_{2}}{C_{1}}+3\frac{C_{3}}{C_{2}}+.....+19\frac{C_{19}}{C_{18}}+20\frac{C_{20}}{C_{19}}$ is equal to (where $C_{r}$ represents $\_{}^{n}C_{r}^{}$ )

• A. $120$
• B. $210$
• C. $180$
• D. $240$

Answer 1

Answer: (B)

$\frac{C_{r}}{C_{r - 1}}=\frac{n - r + 1}{r}=\frac{20 - r + 1}{r}=\frac{21 - r}{r}$
$\Rightarrow r\cdot \frac{C_{r}}{C_{r - 1}}=\left(21 - r\right)$
$\Rightarrow \displaystyle \sum _{r = 1}^{20} r \cdot \frac{C_{r}}{C_{r - 1}}=\displaystyle \sum _{r = 1}^{20} \left(21 - r\right)$
$=21\displaystyle \sum _{r = 1}^{20} 1-\displaystyle \sum _{r = 1}^{20} r$
$=21\times 20-\frac{20 \times 21}{2}$
$=210$