Question

again, telescopic series :

Answer 1

\[\frac{ 1 }{ 1010 . 2016} + \frac{ 1 }{ 1012 . 2014} + \frac{ 1 }{ 1014 . 2012} + ... + \frac{ 1 }{ 2016 . 1010} = ... ? \]

Answer 2

the dot there means times right?

Answer 3

yeah, that times...

Answer 4

for this question really no idea to me

Answer 5

well I see we can write the 1010, 1012,1014, ...
as 1010+2n

Answer 6

where that n=0,1,2,...,506

Answer 7

to figure out the 506
i just solved 1010+2n=2016

Answer 8

http://prntscr.com/4ssja9

Answer 9

can we start n = 1 ? i usually use that :)

Answer 10

now the 2016, 2014, 2012,...,1010
as 2016-2n

Answer 11

\[\sum_{i=0}^{506}\frac{1}{(2n+1010)(2016-2n)}\]
then use partial fractions

Answer 12

yeah ikram, i saw the the series has 252 pairs so just double them

Answer 13

if you want to yes you can write it from i=1 to i=507
just write 2(n-1)+1010
and write 2016-2(n-1)

Answer 14

nice :) the top bound should be 503 i think...

Answer 15

1010+2n=2016
2n=2016-1010
2n=1006
n=503
oh darn

Answer 16

i can't solve algebraic equation

Answer 17

lol doesn't matter the final answer is some weird tiny number

Answer 18

ha ? i i use arithmetic formula i got :
1010,1012, 1014, ..., 2016
2016 = 1010 + (n-1)2
2016 - 1010 = (n-1)2
1006 = (n-1)2
n - 1 = 503
n = 504
is tht wrong ??

Answer 19

well you are using the formula for n=1 to n=504
we did (or i did ) n=0 to n=503

Answer 20

yours is correct too if you choose to use 1/(1010+(n-1)*2)(2016-2(n-1))
from n=1 to n=504

Answer 21

oh ok, that same :)

Answer 22

well, if i start from n = 0
1/(2n + 1010)(2016-2n)
= 1/2(1/2) 1/(n+505)(1008-n)
= 1/4 * 1/(n+505)(1008-n)

partial fractions :
1/(n+505)(1008-n) = A/(n+505) + B/(1008-n)
i got A = 1/1513 and B = A = 1/1513
now i have the series :
1/4 * (1/1513)* (1/(n+505) + 1/(1008-n))
= 1/6052 * (1/(n+505) + 1/(1008-n))

what's next, @myininaya , @ganeshie8 ?

Answer 23

yeah im not getting ideas, i have posted it in MSE.. monitor below thread :
http://math.stackexchange.com/questions/957406/dfrac-1-1010-times-2016-dfrac-1-1012-times-2014-dfrac-1

Answer 24

can we integrate ?

Answer 25

nvm lol

Answer 26

program it xD

Answer 27

\[\frac{1}{4(1513)}[(\frac{1}{505}+\frac{1}{1008})+(\frac{1}{506}+\frac{1}{1007})+(\frac{1}{507}+\frac{1}{1006})+\cdots \\+(\frac{1}{1006}+\frac{1}{507}) +(\frac{1}{1007}+\frac{1}{506}) +(\frac{1}{1008}+\frac{1}{505}) \\ = \frac{1}{4(1513)}[\frac{2}{505}+\frac{2}{506}+\frac{2}{507}+\cdots +\frac{2}{2008}] \\ =\frac{1}{2(1513)}[\frac{1}{505}+\frac{1}{506}+\frac{1}{507}+ \cdots +\frac{1}{2008}]\]

Answer 28

i was thinking it would work doing this oh partial fraction thing since it said it was a telescoping series

Answer 29

i think i got this one :O

Answer 30

@ganeshie8 looks like some people replied to you

Answer 31

yeah still going through them, it seems it won't simplify nicely.. have to approximate only i guess