Question

The value of.......................?
A) 1/3*5+1/5*7+1/7*9+.........+1/23*25 is...
Tell me how to solve problem like this..

Answer 1

I think this sum can be simplified by using partial fractions, e.g. each term is of the form\[\frac{1}{k(k+2)}=\frac{1}{2}\large(\frac{1}{k}-\frac{1}{k+2}\large)\]so we could get a sum where most terms cancel out

Answer 2

hmm, i was trying to think how telescopin gwould fit ...

Answer 3

the thoughts kept going sideways tho

Answer 4

quite difficult dude

Answer 5

the telescoping simplifies it nicely

Answer 6

k

Answer 7

@AliBelly have you used partial fractions and telescoping series?

Answer 8

nope, googling it

Answer 9

\[\frac{ 1 }{ (2k+1)(2k+3) }\]

Answer 10

this will be your series .....now follow the amitsre ' s approch

Answer 11

sorry asnaseer's approch

Answer 12

@AliBelly This might be helpful to you: http://www.mathsisfun.com/algebra/partial-fractions.html

Answer 13

@gorv expression is the better one to use since all terms have odd numbers. The one I gave only applies for k=3,5,7,...

whereas his applies for k=1,2,3,...

Answer 14

yeah i know ..so just for him to take a start tell the equation applicable for his Q

Answer 15

then i said to follow your approch

Answer 16

\[a _{k}=\frac{ 1 }{ \left( 2k+1 \right)\left( 2k+3 \right) },k=1,2,...,11\]
\[\frac{ 1 }{ \left( 2k+1 \right) \left( 2k+3 \right)}=\frac{ 1 }{ \left( 2k+1 \right)\left( -1+3 \right)}+\frac{ 1 }{ \left( -3+1 \right)\left( 2k+3 \right) }\]
\[=\frac{ 1 }{ 2 }\left[ \frac{ 1 }{ 2k+1 }-\frac{ 1 }{ 2k+3 } \right]\]
put k=1,2,...,11 and add vertically downwards