Write the sum to n terms of the series 11-2-12-3-13-4-1 nn-1

Sn=11.2+12.3+13.4+...+1n #160;(n+1) nbsp; nbsp; nbsp; =11 12+12 13+13 14+...+1n 1 #160;(n+1)=1 1n+1=nn+1 ...

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Quantitative Aptitude

Series

Write the sum...

Question

Write the sum to n terms of the series $\frac{1}{1.2} + \frac{1}{2.3} + \frac{1}{3.4} + . . . + \frac{1}{n (n + 1)}$ .

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Solution

$S_{n} = \frac{1}{1.2} + \frac{1}{2.3} + \frac{1}{3.4} + . . . + \frac{1}{n (n + 1)}$
$= (\frac{1}{1} - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + (\frac{1}{3} - \frac{1}{4}) + . . . + (\frac{1}{n} - \frac{1}{(n + 1)}) = 1 - \frac{1}{(n + 1)} = \frac{n}{n + 1}$

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Write the sum to n terms of the series 11.2+12.3+13.4+...+1n (n+1).

Sn=11.2+12.3+13.4+...+1n (n+1) =11-12+12-13+13-14+...+1n-1 (n+1)=1-1n+1=nn+1

Write the sum to n terms of the series $\frac{1}{1.2} + \frac{1}{2.3} + \frac{1}{3.4} + . . . + \frac{1}{n (n + 1)}$ .

$S_{n} = \frac{1}{1.2} + \frac{1}{2.3} + \frac{1}{3.4} + . . . + \frac{1}{n (n + 1)}$
$= (\frac{1}{1} - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + (\frac{1}{3} - \frac{1}{4}) + . . . + (\frac{1}{n} - \frac{1}{(n + 1)}) = 1 - \frac{1}{(n + 1)} = \frac{n}{n + 1}$