If (1+x)n=C0+C1x+C2x2+.......+Cnxn, then C1C0+2C2C1+3C3C2+........+nCnCn−

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If

(1 + x)^{n} = C_{0} + C_{1} x + C_{2} x^{2} + . . . . . . . + C_{n} x^{n}

, then

\frac{C_{1}}{C_{0}} + \frac{2 C_{2}}{C_{1}} + \frac{3 C_{3}}{C_{2}} + . . . . . . . . + \frac{n C_{n}}{C_{n - 1}} =

Options:

A . n(n−1)2

B . n(n+2)2

C . n(n+1)2

D . (n−1)(n−2)2

Answer: Option C
:
C

\frac{C_{1}}{C_{0}} + 2 \cdot \frac{C_{2}}{C_{1}} + 3 \cdot \frac{C_{3}}{C_{2}} + . . . . . . . . + n \cdot \frac{C_{n}}{C_{n - 1}}

= \frac{n}{1} + 2 \frac{\frac{n (n - 1)}{1.2}}{n} + 3 \frac{\frac{n (n - 1) (n - 2)}{3.2.1}}{\frac{n (n - 1)}{1.2}} + . . . . . . + n . \frac{1}{n}

= n + (n - 1) + (n - 2) . . . . . . . . + 1 = \sum n = \frac{n (n + 1)}{2}

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