Name: Sum of Infinite Terms of G.P.
Uploaded: 2019-07-19
Duration: 3 min 36 s
Description: Do you know that we can find the sum of an infinite number of terms of G.P.? Of course, there will be some conditions. Check this video to learn more.

Question

Find sum of the series

\frac{n}{1 \cdot 2 \cdot 3} + \frac{n - 1}{2 \cdot 3 \cdot 4} + \frac{n - 2}{3 \cdot 4 \cdot 5} + \dots . .

up to

n

terms.

Accepted Answer

Given series $\frac{n}{1 \cdot 2 \cdot 3} + \frac{n - 1}{2 \cdot 3 \cdot 4} + \frac{n - 2}{3 \cdot 4 \cdot 5} + \dots upto n terms$

General term of the series can be written as

$T_{r} = \frac{n + 1 - r}{r (r + 1) (r + 2)}$

$T_{r} = \frac{n + 1}{r (r + 1) (r + 2)} - \frac{1}{(r + 1) (r + 2)}$

Sum of the series $S = \sum_{r = 1}^{n} T_{r}$

$S = (n + 1) \sum_{r = 1}^{n} \frac{1}{r (r + 1) (r + 2)} - \sum_{r = 1}^{n} \frac{1}{(r + 1) (r + 2)}$

$\Rightarrow S = \frac{(n + 1)}{2} \sum_{r = 1}^{n} \frac{1}{(r + 1)} [\frac{1}{r} - \frac{1}{r + 2}] - \sum_{r = 1}^{n} (\frac{1}{r + 1} - \frac{1}{r + 2})$

$\Rightarrow S = \frac{(n + 1)}{2} [\frac{1}{1 \cdot 2} - \frac{1}{2 \cdot 3} + \frac{1}{2 \cdot 3} - \frac{1}{3 \cdot 4} + \frac{1}{4 \cdot 3} - \frac{1}{4 \cdot 5} \dots + \frac{1}{n (n + 1)} - \frac{1}{(n + 1) (n + 2)}] - [\frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + \frac{1}{4} - \frac{1}{5} + \frac{1}{5} - \frac{1}{6} + \dots . + \frac{1}{(n + 1)} - \frac{1}{(n + 2)}]$

$\Rightarrow S = \frac{n + 1}{2} [\frac{1}{2} - \frac{1}{(n + 1) (n + 2)}] - [\frac{1}{2} - \frac{1}{(n + 2)}]$

$\Rightarrow S = \frac{n + 1}{2} [\frac{1}{2} - \frac{1}{(n + 1) (n + 2)}] - \frac{1}{2} + \frac{1}{(n + 2)}$

$\Rightarrow S = \frac{n + 1}{4} - \frac{1}{2 (n + 2)} - \frac{1}{2} + \frac{1}{(n + 2)}$

$\Rightarrow S = \frac{n}{4} + \frac{1}{4} - \frac{1}{2} + \frac{1}{2 (n + 2)}$

$\Rightarrow S = \frac{n}{4} + \frac{1}{2 (n + 2)} - \frac{1}{4}$

$S = \frac{n (n + 1)}{4 (n + 2)}$ Answer.

Find sum of the series $\frac{n}{1 \cdot 2 \cdot 3} + \frac{n - 1}{2 \cdot 3 \cdot 4} + \frac{n - 2}{3 \cdot 4 \cdot 5} + \dots . .$ up to $n$ terms.

Important Questions on Sequences and Series

Learn and Prepare for any exam you want !

Find sum of the series n1·2·3+n-12·3·4+n-23·4·5+….. up to n terms.

Important Questions on Sequences and Series

Learn and Prepare for any exam you want !

Find sum of the series $\frac{n}{1 \cdot 2 \cdot 3} + \frac{n - 1}{2 \cdot 3 \cdot 4} + \frac{n - 2}{3 \cdot 4 \cdot 5} + \dots . .$ up to $n$ terms.