If Cr=nCr, then value of 2C1C0+2C2C1+3C3C2+………+nCnCn−1 is

If $C_{r} =^{n} C_{r},$ then value of $2 (\frac{C_{1}}{C_{0}} + 2 \frac{C_{2}}{C_{1}} + 3 \frac{C_{3}}{C_{2}} + . . . . . . . . . + n \frac{C_{n}}{C_{n - 1}})$ is

A
$n (n - 1)$
B
$n (n + 1)$
C
$n^{2} - 1$
D
$n^{2} + 1$

Solution:

$\begin{array}{l} \frac{C_{r}}{C_{r - 1}} = \frac{n!}{r! (n - r)!} \cdot \frac{(r - 1)! (n - r + 1)!}{n!} \\ = \frac{n - r + 1}{r} \\ \Rightarrow r \frac{C_{r}}{C_{r - 1}} = n - r + 1 \\ \Rightarrow 2 \sum_{r = 1}^{n} r \frac{C_{r}}{C_{r - 1}} = 2 \sum_{r = 1}^{n} (n - r + 1) = n (n + 1) \end{array}$

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