If 1-xn-C0-C1 x-C2 x2- -Cn xn- then C1-C0-2 C2-C1-3 C3-C2- - - nC n-Cn-1-A- nn-1-2B- nn-2-2C- n n-1-2D- n -1 n -2-2

The correct option is C n(n+1)/2 C_1/C_0 + 2 ·C_2/C_1+ 3 · nbsp; C_3/C_2+ ........+n ·C_n/C_n 1 nbsp;=n/1 nbsp;+ 2 n(n 1)/1.2/n nbsp;+ 3 n(n 1)(n 2)/3 ...

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Byju's Answer

Standard XI

Mathematics

Binomial Coefficients

If 1+xn=C0+C1...

Question

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

$\frac{n (n - 1)}{2}$

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$\frac{n (n + 2)}{2}$

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$\frac{n (n + 1)}{2}$

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$\frac{(n - 1) (n - 2)}{2}$

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Solution

The correct option is C
$\frac{n (n + 1)}{2}$

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

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

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

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