Question

Find the sum to n terms \[\frac{ 1 }{ 1*2}+\frac{ 1 }{ 2*3}+\frac{ 1 }{ 3*4 }+.......\]

Answer 1

break each term to fractions

Answer 2

1/2 + 1/6+1/12.......

Answer 3

no, write 1 as 2-1 above 2*1
and 3-2 above 3*2.....and so on, then break into fractions.....

Answer 4

then..)

Answer 5

@satellite73

Answer 6

did u even try?

Answer 7

Yup...i didnt find any thing..there

Answer 8

1-1/2 + 1/2 -1/3 + 1/3 - 1/4 + ....
didn't u get this ?

Answer 9

I did this in another way using kth order difference and got 1/(n+n^2)

Answer 10

n^2 = n(n+1)(2n+1)/6

n=n(n+1)/2

Answer 11

you have
summation (1/n(n+1) )

now,, 1/n(n+1) = 1/n - 1/(n+1)

hope you get it ..

Answer 12

cant we take it as 1/(n^2+n)

Answer 13

you can take it any how you want,,but what ever you take ,,you should be comfortable with that,,
tell me,,what progress you make with n^2 + n thing?

Answer 14

n^2 = n(n+1)(2n+1)/6

Answer 15

n=n(n+1)/2

Answer 16

you dont have n^2 .. you have 1/n^2
and not only that,,its 1/(n^2 +n) ,,you're confusing yourself,,re-think please..

Answer 17

Why..

Answer 18

1/n^2+n = 1/(n(2n+1)(n+1)/6 +n(n+1)/2

Answer 19

you want to say
summation 1/n = 2/n(n+1) ??

Answer 20

Yes..)why is it wrong

Answer 21

try this one for n=1 ,2 etc and see for yourself..

Answer 22

and logically,,
summation 1/n = 1/1 + 1/2 + 1/3 .......
and
summation n = 1 + 2+ 3+ ......

see it ..is one of them reciprocal of the other ? o.O

Answer 23

yes..u r correct...)....Continue by ur method

Answer 24

i already hinted you,,you should try and continue yourself..