Sum of $\frac{1}{1\times3\times5}+\frac{1}{3\times5\times7}+\cdots + \frac{1}{13\times15\times17}$

Question

I solved like this, but it took me lots of time and I have to use the calculator. But in exam I can't use the calculator and I need to be fast.

(1) Is there any alternative to solve this easily?

(2) How to solve this if given for $n$ terms?

My work: \begin{align} \frac{1}{1\times3\times5}&+\frac{1}{3\times5\times7}+\cdots +\frac{1}{13\times15\times17}\\ &=\frac{1}{1\times3\times5}+\frac{1}{3\times5\times7}+\frac{1}{5\times7\times9}+\frac{1}{7\times9\times11}+\frac{1}{9\times11\times13}\\ &\ \ \ \ \ +\frac{1}{11\times13\times15}+\frac{1}{13\times15\times17}\\ &=\frac{1}{15}[1+1/7+1/21]+\frac{1}{11\times9}[1/7+1/13]+\frac{1}{15\times17}[1/11+1/17]\\ &=7/85 \end{align}

@Typhon what do you mean? if anything confusing please edit it — Rohit, Nov 01 '17 at 04:34
That is a telescopic sum and similar questions have already appeared here multiple times - https://math.stackexchange.com/questions/1260728/find-the-summation-of-the-series-sum-n-15-frac12n-12n12n3 — Jack D'Aurizio, Nov 01 '17 at 16:24

Khosrotash · Accepted Answer · 2017-11-01T05:03:31.930

5

$$\quad{\frac{1}{1\times3\times5}+\frac{1}{3\times5\times7}+…\frac{1}{13\times15\times17}=\\\sum_{n=1}^{7}\frac{1}{(2n-1)(2n+1)(2n+3)}=\\ \sum_{n=1}^{7}\frac{1}{4}(\frac{1}{(2n-1)(2n+1)}-\frac{1}{(2n+1)(2n+3)})=\\\frac14(\frac{1}{1.3}-\frac{1}{15.17})}$$ remark:$$\frac{1}{(2n-1)(2n+1)(2n+3)}=\\\frac 44 .\frac{1}{(2n-1)(2n+1)(2n+3)}=\\ \frac1 4\frac{4}{(2n-1)(2n+1)(2n+3)}=\\ \frac1 4\frac{(2n+3)-(2n-1)}{(2n-1)(2n+1)(2n+3)}=\\ \frac1 4(\frac{(2n+3)}{(2n-1)(2n+1)(2n+3)}-\frac{(2n-1)}{(2n-1)(2n+1)(2n+3)})=\\$$ remark 2: $$\quad{\frac{1/8}{(2n-1)}-\frac{1/4}{2n+1}+\frac{1/8}{2n+3}=\\ \frac{1/8}{(2n-1)}-\frac{1/8+1/8}{2n+1}+\frac{1/8}{2n+3}=\\ \frac 18(\frac{1}{(2n-1)}-\frac{1+1}{(2n+1)}+\frac{1}{(2n+3)})=\\ \frac 18(\frac{1}{(2n-1)}-\frac{1}{(2n+1)}-\frac{1}{(2n+1)}+\frac{1}{(2n+3)})=\\\frac18 (\frac{2}{(2n-1)(2n+1)}-\frac{2}{(2n+1)(2n+3)})=\\ \frac{1}{4}(\frac{1}{(2n-1)(2n+1)}-\frac{1}{(2n+1)(2n+3)})=\\ \frac{1}{4}(\frac{(2n+3)-(2n-1)}{(2n-1)(2n+1) (2n+3)})=\\\frac{4}{4}.\frac{1}{(2n-1)(2n+1) (2n+3)}}$$

edited Nov 01 '17 at 05:03

answered Nov 01 '17 at 04:23

Khosrotash

24,922

how yo did the 3rd step? – Rohit Nov 01 '17 at 04:32
@Rohit This is a kind of partial fraction decomposition. Are you familiar with that technique? – Carl Schildkraut Nov 01 '17 at 04:40
@Rohit : I fixed it ,and show the method step by step – Khosrotash Nov 01 '17 at 04:42
@Khosrotash but i am getting partial fraction like this $\frac{1}{(2n-1)(2n+1)(2n+3)}$=$\frac{1/8}{(2n-1)}+\frac{1/4}{2n+1}+\frac{1/8}{2n+3}$ – Rohit Nov 01 '17 at 04:46
@Rohit :it can be converted to that form , you need some calculation – Khosrotash Nov 01 '17 at 04:48
@Khosrotash YES – Rohit Nov 01 '17 at 04:50
@Rohit :please look again , I put the steps – Khosrotash Nov 01 '17 at 05:04

lab bhattacharjee · Answer 2 · 2017-11-01T05:27:11.337

2

Hint:

$$\dfrac4{(2n-1)(2n+1)(2n+3)}$$

$$=\dfrac{2n+3-(2n-1)}{(2n-1)(2n+1)(2n+3)}=\dfrac1{(2n-1)(2n+1)}-\dfrac1{(2n+1)(2n+3)}=f(n)-f(n+1)$$

where $$f(m)=\dfrac1{(2m-1)(2m+1)}$$

See Telescoping series

edited Nov 01 '17 at 05:27

answered Nov 01 '17 at 04:21

lab bhattacharjee

274,582

sorrry I used that too and I stuck there where you are giving me hint,can you help me how to procedd it further? – Rohit Nov 01 '17 at 04:22
@Rohit, if $$f(n)=1/(2n-1)(2n+1),$$ $$f(n+1)=?$$ – lab bhattacharjee Nov 01 '17 at 04:25
its $1/(2n+1)(2n+3)$ but what now,not getting you? – Rohit Nov 01 '17 at 04:28
@Rohit, so, we have $$f(n)-f(n+1)$$ right? Now set $n=1,2,\cdots ,7$ – lab bhattacharjee Nov 01 '17 at 04:30
1

@Rohit, See also : https://math.stackexchange.com/questions/560816/find-the-sum-of-the-series-sum-frac1nn1n2/560897#560897 – lab bhattacharjee Nov 01 '17 at 05:28
Thankyou so much sir I upvoted your answer it helped....However I am accepting Khosrotash's answer as you have sea of reputation :-)) – Rohit Nov 01 '17 at 05:33
@Rohit, upvote/reputation doesn't matter as long as answer is understandable to u – lab bhattacharjee Nov 01 '17 at 05:42
Sir I understood you answer also reading the wikipedia link and another problem link you gave me I understood it completely. :-) thanks again – Rohit Nov 01 '17 at 06:02

Sum of $\frac{1}{1\times3\times5}+\frac{1}{3\times5\times7}+\cdots + \frac{1}{13\times15\times17}$

2 Answers2