JEE Mains Sequence and Series 1

Calculus Level 5

If S S denotes the sum of infinite terms of the series 1 1 × 3 × 7 + 1 3 × 5 × 9 + 1 5 × 7 × 11 + ⋯ \dfrac{1}{1\times 3\times 7}+\dfrac{1}{3\times 5\times 9}+\dfrac{1}{5\times 7\times 11}+\cdots , then calculate 180 S 180S

This problem is a part of My Picks for JEE Mains 1


The answer is 11.00.

Pranjal Jain by Pranjal Jain , 23, India

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2 solutions

The terms are of the form 1 ( 2 k − 1 ) ( 2 k + 1 ) ( 2 k + 5 ) \dfrac{1}{(2k - 1)(2k + 1)(2k + 5)} for positive integers k k .

Now the partial fraction decomposition of these terms is

1 12 2 k − 1 − 1 8 2 k + 1 + 1 24 2 k + 5 . \dfrac{\frac{1}{12}}{2k - 1} - \dfrac{\frac{1}{8}}{2k + 1} + \dfrac{\frac{1}{24}}{2k + 5}.

We can thus write S S as

1 12 ( 1 + 1 3 + 1 5 + 1 7 + 1 9 + . . . . ) − \dfrac{1}{12}\left (1 + \dfrac{1}{3} + \dfrac{1}{5} + \dfrac{1}{7} + \dfrac{1}{9} + ....\right ) -

1 8 ( 1 3 + 1 5 + 1 7 + 1 9 + . . . . ) + \dfrac{1}{8}\left (\dfrac{1}{3} + \dfrac{1}{5} + \dfrac{1}{7} + \dfrac{1}{9} + ....\right ) +

1 24 ( 1 7 + 1 9 + . . . . . . ) = \dfrac{1}{24}\left (\dfrac{1}{7} + \dfrac{1}{9} + ......\right ) =

1 12 + ( 1 12 − 1 8 ) ( 1 3 + 1 5 ) + ( 1 12 − 1 8 + 1 24 ) ( 1 7 + 1 9 + . . . . . ) = \dfrac{1}{12} + \left (\dfrac{1}{12} - \dfrac{1}{8}\right )\left (\dfrac{1}{3} + \dfrac{1}{5}\right ) +\left (\dfrac{1}{12} - \dfrac{1}{8} + \dfrac{1}{24}\right )\left (\dfrac{1}{7} + \dfrac{1}{9} + .....\right ) =

1 12 − ( 1 24 ) ( 8 15 ) + 0 = 22 360 \dfrac{1}{12} - \left (\dfrac{1}{24}\right )\left (\dfrac{8}{15}\right ) + 0 = \dfrac{22}{360} ,

which when multiplied by 180 180 yields 22 2 = 11 \dfrac{22}{2} = \boxed{11} .

23 Helpful 0 Interesting 0 Brilliant 0 Confused

How did you get the partial fraction decomposition of these terms? By the way, nice solution! Upvoted :)

Yogesh Verma - 6 years, 3 months ago

Yeah! I also did the same way that u did.

Priyesh Pandey - 6 years, 3 months ago

Exactly ! A nice solution to a nice question . :D

Keshav Tiwari - 6 years, 2 months ago

All the level 4 problems on sequence and series that I have solved so far here seem to be not harder than a level 2 problem. By the way, before finding the series sum you must prove that the series is convergent, that is, its infinite series sum exists. In this case proving the convergence is trivial since we can very easily apply the ratio test.

Kuldeep Guha Mazumder - 5 years, 6 months ago

9 Helpful 2 Interesting 1 Brilliant 0 Confused

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