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AP EAMCET Engineering 26 Apr 2017 Shift 1 Paper

Section: Mathematics

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Question : 3 of 160

Marks: +1, -0

The sum to first n terms of

\frac{1}{2\cdot5} + \frac{1}{5\cdot8} + \frac{1}{8\cdot11} + \dots

is

$\frac{3n}{2(3n+2)}$
$\frac{}{3n+2}$
$\frac{n}{2(3n+2)}$
${n}/{2(3n+2)}$

Solution:

Let

S_n = \frac{1}{2\cdot5} + \frac{1}{5\cdot8} + \frac{1}{8\cdot11} + \dots

n

terms

It is simplified as,

S_n = \frac{1}{3} \left[ \frac{3}{2\cdot5} + \frac{8-5}{5 \cdot 8} + \frac{11-8}{8 \cdot 11} + \dots + n \text{ terms} \right]

= \frac{1}{3} \left[ \frac{5-2}{2\cdot5} + \frac{3}{5 \cdot 8} + \frac{3}{8 \cdot 11} + \dots + n \text{ terms} \right]

= \frac{1}{3} \left[ \left( \frac{1}{2} - \frac{1}{5} \right) + \left( \frac{1}{5} - \frac{1}{8} \right) + \left( \frac{1}{8} - \frac{1}{11} \right) \right]

+ \dots + \left( \frac{1}{3n-1} - \frac{1}{3n+2} \right) ]

= \frac{1}{3} \left[ \frac{1}{2} - \frac{1}{3n+2} \right]

Solve further,

S_n = \frac{1}{3} \left[ \frac{3n+2-2}{2(3n+2)} \right]

= \frac{n}{2(3n+2)}

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