In the expansion of (1 + x)n ,C1C0+2C2C1+3C3C2+…+nCnCn−1 is equal to

A
$\frac{(n + 1)}{2}$
B
$\frac{n}{2}$
C
$\frac{n (n + 1)}{2}$
D
None of these

Solution:

Since, we know that $\frac{^{n} C_{r}}{^{n} C_{r - 1}} = \frac{n - r + 1}{r}$

$∴ \frac{C_{1}}{C_{0}} = n, \frac{C_{2}}{C_{1}} = \frac{n - 1}{2}, \frac{C_{3}}{C_{2}} = \frac{n - 2}{3}, \dots$

$Thus, \frac{C_{1}}{C_{0}} + 2 \frac{C_{2}}{C_{1}} + 3 \frac{C_{3}}{C_{2}} + \dots = n + 2 \cdot \frac{n - 1}{2}$ $+ 3 \cdot \frac{n - 2}{3} + \dots + n \cdot \frac{1}{n}$

$\begin{array}{l} = [n + (n - 1) + (n - 2) + \dots + 1] \\ = \frac{n (n + 1)}{2} \end{array}$

In the expansion of (1 + x)ⁿ , $\frac{C_{1}}{C_{0}} + 2 \frac{C_{2}}{C_{1}} + 3 \frac{C_{3}}{C_{2}} + \dots + n \frac{C_{n}}{C_{n - 1}}$ is equal to

Solution:

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