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  4. In the expansion of (1 + x)n, (C1/C0)+2(C2/C1)+3(C3/C2)+ ldots +n (Cn/Cn-1) is equal to

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

Binomial Theorem

Solution:

Since, we know-that $\frac{^{n}C_{r}}{^{n}C_{r-1}}=\frac{n-r+1}{r}$
$\therefore \frac{C_{1}}{C_{0}} = n$, $\frac{C_{2}}{C_{1}} = \frac{n-1}{2}$, $\frac{C_{3}}{C_{2}}= \frac{n-2}{3}$, $\ldots$
Thus, $\frac{C_{1}}{C_{0}}+2\frac{C_{2}}{C_{1}}+3\frac{C_{3}}{C_{2}}+\ldots +n \frac{C_{n}}{C_{n-1}}$
$= n+2\cdot \frac{n-1}{1}+3\cdot\frac{n-2}{3}+\ldots+n\cdot\frac{1}{n}$
$= \left[n + \left(n - 1\right) + \left(n - 2\right) + ... + 1\right] =\frac{n\left(n+1\right)}{2}$