ind the sum of n terms of :- 1/2*5+1/5*8+1/8*11+ - Maths - Sequences and Series

Question

ind the sum of n terms of :- 1/2*5+1/5*8+1/8*11+

Chetna Khanna · Accepted Answer

$\frac{1}{2 \times 5} + \frac{1}{5 \times 8} + \frac{1}{8 \times 11} + ....$

Let S_n be the sum of n terms of this series.

$S_{n} = \frac{2}{2 \times 5} + \frac{1}{5 \times 8} + \frac{1}{8 \times 11} + .... + \frac{1}{(3 n - 4) (3 n - 1)} = \frac{3}{3} [\frac{1}{2 \times 5} + \frac{1}{5 \times 8} + \frac{1}{8 \times 11} + .... + \frac{1}{(3 n - 4) (3 n - 1)}] = \frac{1}{3} [\frac{3}{2 \times 5} + \frac{3}{5 \times 8} + \frac{3}{8 \times 11} + .... + \frac{1}{(3 n - 4) (3 n - 1)}] = \frac{1}{3} [\frac{(5 - 2)}{2 \times 5} + \frac{(8 - 5)}{5 \times 8} + \frac{(11 - 8)}{8 \times 11} + .... + \frac{(3 n - 1) - (3 n - 4)}{(3 n - 4) (3 n - 1)}] = \frac{1}{3} [\frac{5}{2 \times 5} - \frac{2}{2 \times 5} + \frac{8}{5 \times 8} - \frac{5}{5 \times 8} + \frac{11}{8 \times 11} - \frac{8}{8 \times 11} + .... + \frac{(3 n - 1)}{(3 n - 4) (3 n - 1)} - \frac{(3 n - 4)}{(3 n - 4) (3 n - 1)}]$

$= \frac{1}{3} [\frac{1}{2} - \frac{1}{5} + \frac{1}{5} - \frac{1}{8} + \frac{1}{8} - \frac{1}{11} + .... + \frac{1}{(3 n - 4)} - \frac{1}{(3 n - 1)}] = \frac{1}{3} [\frac{1}{2} - \frac{1}{(3 n - 1)}] = \frac{1}{3} [\frac{3 n - 1 - 2}{2 (3 n - 1)}] = \frac{1}{3} [\frac{3 n - 3}{2 (3 n - 1)}] = \frac{1}{3} \times \frac{3 (n - 1)}{2 (3 n - 1)} = \frac{(n - 1)}{2 (3 n - 1)}$

Note: Put the values of n to get the sum of n terms.

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