The sum of the series \; 1/2 · 3 + \; 1/4 · 5 + \; 1/6 · 7 + is

Correct Answer is : log⁡(e2) ≤ft( e/2 )log(2e)

Correct Answer is : log⁡(e2) ≤ft( e/2 )log(2e)

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CGPET 2016 Solved Paper

Section: Mathematics

Question : 103 of 150

Marks: +1, -0

The sum of the series

\; \frac{1}{2 \cdot 3} + \; \frac{1}{4 \cdot 5} + \; \frac{1}{6 \cdot 7} + \dots

Solution:

Given series is

\frac{1}{2.3} + \frac{1}{4.5} + \frac{1}{6.7}

∴

a_n = \frac{1}{( \text{nth term of } 2,4,6 \dots)( \text{nth term of } 3,5,7 \dots)}

= \frac{1}{(2n)(2n+1)}

= \left( \frac{1}{2n} - \frac{1}{2n+1} \right)

[by using partial fraction]

Thus,

a_k = \left( \frac{1}{2n} - \frac{1}{2n+1} \right)

On putting

k = 1,2,3, \dots \text{ we get } a_1 = \left( \frac{1}{2} - \frac{1}{3} \right)

a_2 = \left( \frac{1}{4} - \frac{1}{5} \right)

a_3 \left( \frac{1}{6} - \frac{1}{7} \right)

a_n = \left( \frac{1}{2n} - \frac{1}{2n+1} \right)

∴

\sum\limits_{i=1}^{n} a_k

= \left( \frac{1}{2} - \frac{1}{3} \right) + \left( \frac{1}{4} - \frac{1}{5} \right) + \left( \frac{1}{6} - \frac{1}{7} \right) + \dots + \left( \frac{1}{2n} - \frac{1}{2n+1} \right)

= - \left( - \frac{1}{2} + \frac{1}{3} + - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \frac{1}{7} + \dots \right)

= 1 - \left( 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \frac{1}{7} + \dots \right)

= \log e - \log 2 = \log \left( \frac{e}{2} \right)

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