Find the sum to n terms:
1\1*3 + 2\1*3*5 + 3\1*3*5*7......to n terms.
1Tn = n1.2.3...............(2n+1)
= 12 ( 2n+1-11.2.3............(2n+1) )
= 12( 11.2.3.....(2n-1) - 11.2.3.........(2n+1) )
= 12 (tn-1 - tn)
where tn = 11.2.3........(2n+1)
Now put n=2,3,4.and so on upto n terms. And add. U will find that the terms are cancelling out diagonally until the first and last terms remain.
Sn = 12 (t1 - tn )
= 12 (1 - 11.2.3.......(2n+1) )
I can't solve after this.
Pls reply if my answer is correct.
1dude, u hv taken Tn incorrect.
tn = n / 1.3.5.....(2n+1) = 1/2 [ (2n+1)-1 / 1.3.5....(2n+1) ]
= 1/2 . [ 1 / 1.3.5.....(2n-1) - 1 / 1.3.5.....(2n+1) ] = 1/2 [ Tn-1 - Tn ]
where Tn = 1 / 1.3.5.....(2n+1)
now,continue with the summation method by putting n=1,2,3,...
finally,u would get : 2 (Sn - t1) = T1 - Tn.
put the values. u'll get:
Sn = 1/2 [ 1 - {1 / 1.3.5.....(2n=1)} ].
i did d same thing what u did,just write tn properly.
1Wow, Thank you both for the answers . I was stuck on it for quite sometime.