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Sum of Infinite Terms of a GP

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Question

Find the sum of the series when n = 10. $\frac{C_{1}}{C_{0}}$ + 2 $\frac{C_{2}}{C_{1}}$ + 3 $\frac{C_{3}}{C_{2}}$ + ...............+ n $\frac{C_{n}}{C_{r - 1}}$

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Solution

When we observe the terms they are of the form

r × $\frac{^{n} C_{r}}{^{n} C_{r - 1}}$

We know $\frac{^{n} C_{r}}{^{n} C_{r - 1}}$ = $\frac{n - r + 1}{r}$

⇒ r × $\frac{^{n} C_{r}}{^{n} C_{r - 1}}$ = n - r + 1

⇒ $\frac{C_{1}}{C_{0}}$ + 2 $\frac{C_{2}}{C_{1}}$ + 3 $\frac{C_{3}}{C_{2}}$ .................. $n \sum r = 1 r$ $\frac{^{n} C_{r}}{^{n} C_{r - 1}}$

= $n \sum r = 1 n - r + 1$

= $n \sum r = 1 n$ - $n \sum r = 1 r$ + $n \sum r = 1 1$

= n × n - $\frac{n (n + 1)}{2}$ + n

= $n^{2}$ + n - $\frac{(n^{2} + n)}{2}$

= $\frac{2 n^{2} + 2 n - n^{2} - n}{2}$

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

Q. Prove that the sum of the series

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

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Q. The value of the expression

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

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