special series

T_n = 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 (t_n-1 - t_n)

where t_n = 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.

S_n = 12 (t₁ - t_n )

= 12 (1 - 11.2.3.......(2n+1) )

I can't solve after this.

Pls reply if my answer is correct.

dude, u hv taken T_n incorrect.

t_n = 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 [ T_n-1 - T_n ]

where T_n = 1 / 1.3.5.....(2n+1)

now,continue with the summation method by putting n=1,2,3,...

finally,u would get : 2 (S_n - t₁) = T₁ - T_n.

put the values. u'll get:

S_n = 1/2 [ 1 - {1 / 1.3.5.....(2n=1)} ].

i did d same thing what u did,just write t_n properly.