Find the series' sum: 3/1.2.4 + 4/2.3.5 + 5/3.4.6...

[cooldudeachyut]

Question : Find the sum of series :- 3/1.2.4 + 4/2.3.5 + 5/3.4.6 .............. n terms

My attempt :- Since nth term of series is (n+2)/n(n+1)(n+3), I have to find ∑(n+2)/n(n+1)(n+3)
I decomposed it into partial fractions as :-

∑ 2/3n - 1/2(n+1) - 1/6(n+3)

Now I don't know how to proceed at this point.
Please help.

 

[stapel]

Question : Find the sum of series :- 3/1.2.4 + 4/2.3.5 + 5/3.4.6 .............. n terms

I'm not familiar with decimal numbers having TWO decimal points in them. Are you perhaps using the decimal point in place of the "times" symbol? If so, try here to refresh on standard mathematical formatting in text-only environments.

Assuming that the "point" means "times", you have:

. . . . .\(\displaystyle \mbox{Find the sum of the first }\, n\, \mbox{ terms: }\, \dfrac{3}{1\, \cdot\, 2\, \cdot\, 4}\, +\, \dfrac{4}{2\, \cdot\,3\, \cdot\,5}\, +\, \dfrac{5}{3\, \cdot\, 4\, \cdot\, 6}\, +\, ...\)

My attempt :- Since nth term of series is (n+2)/n(n+1)(n+3), I have to find ∑(n+2)/n(n+1)(n+3)
I decomposed it into partial fractions as :-

∑ 2/3n - 1/2(n+1) - 1/6(n+3)

Decomposition is probably a smart idea, and leads to:

. . . . .$\displaystyle \displaystyle \sum_{i\, =\, 1}^n\, \left(\, \dfrac{2}{3i}\, -\, \dfrac{1}{2\, (i\, +\, 1)}\, -\, \dfrac{1}{6\, (i\, +\, 3)}\, \right)$

Now try working with indices. For instance:

. . . . .$\displaystyle \displaystyle \sum_{i\, =\, 1}^n\, \dfrac{-1}{6\, (i\, +\, 3)}\, =\, -\dfrac{1}{6}\, \sum_{i\, =\, 1}^n\, \dfrac{1}{i\, +\, 3}$

Adjusting the indices a bit, we get:

. . . . .$\displaystyle \displaystyle \sum_{i\, =\, 1}^n\, \dfrac{1}{i\, +\, 3}\, =\, \sum_{i\, =\, 4}^{n\, +\, 3}\, \dfrac{1}{i}$

Does this help at all?

(By the way, you can check your results here.)

 

[cooldudeachyut]

. . . . .$\displaystyle \displaystyle \sum_{i\, =\, 1}^n\, \dfrac{1}{i\, +\, 3}\, =\, \sum_{i\, =\, 4}^{n\, +\, 3}\, \dfrac{1}{i}$

How to solve after that?

 

[stapel]

How to solve after that?

What did you get when you adjusted the indices and worked with the overlap? For instance, you followed the provided step and did the same sort of thing with the second of the partial fractions, eventually obtaining:

. . . . .$\displaystyle \displaystyle \sum_{i\, =\,4}^n\, \dfrac{2}{3}\, \cdot\, \dfrac{1}{i}\, +\, \sum_{i\, =\,4}^n\, -\dfrac{1}{2}\, \cdot\, \dfrac{1}{i}\, +\, \sum_{i\, =\,4}^n\, -\dfrac{1}{6}\, \cdot\, \dfrac{1}{i}$

This immediately simplifies, obviously. Then you were left with the remaining terms, which you summed and... then what?

Please be complete. Thank you!