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

# If then value of $2\left(\frac{{C}_{1}}{{C}_{0}}+2\frac{{C}_{2}}{{C}_{1}}+3\frac{{C}_{3}}{{C}_{2}}+.........+n\frac{{C}_{n}}{{C}_{n-1}}\right)$ is

1. A

$n\left(n-1\right)$

2. B

$n\left(n+1\right)$

3. C

${n}^{2}-1$

4. D

${n}^{2}+1$

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### Solution:

$\begin{array}{l}\frac{{C}_{r}}{{C}_{r-1}}=\frac{n!}{r!\left(n-r\right)!}\cdot \frac{\left(r-1\right)!\left(n-r+1\right)!}{n!}\\ =\frac{n-r+1}{r}\\ ⇒r\frac{{C}_{r}}{{C}_{r-1}}=n-r+1\\ ⇒2\sum _{r=1}^{n} r\frac{{C}_{r}}{{C}_{r-1}}=2\sum _{r=1}^{n} \left(n-r+1\right)=n\left(n+1\right)\end{array}$

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