find:

{1/(2n+1)}+{1/3(2n+1)^3}+{1/5(2n+1)^5}+...........to n terms.

This is a series whose reciprocals are in arithmetico geometric series

take 1/2n+1 = Z (say)

then, S =  Z + 3Z³ + 5Z⁵ +.............N TERMS

T_n of this series is = [ Z + (n-1)2] Z^2(n-1)

S_n =  Z             2.Z²       2.Z²ⁿ          [ Z + (n-1) 2 ].Z²ⁿ   

       ------    + -----------  -  --------   -  ------------------------

       (1-Z²)      (1-Z^2⁾²      (1-Z²)²           (1-Z²)

 

on simplyfying i get

 

=  Z ( Z³ + Z²(1-2n) + (2n -1) Z + 1 - 2Z^2(n-1))

    ------------------------------------------------------------

                      ( 1 - Z ^2 ⁾² 

 This is too long but u can substitute for Z and get the ans

But if this comes in objective section it wud be better..........

 

Cheeeeeeers 

Take the given series as

 

S= 1/2n+1 +1/3(2n+1)^3+.....................+1/(2n-1)(2n+1)^(2n-1)     -(1)   {this is the nth therm}

 

Multiply both sides by 1/ (2n+1)²  and subtract this expression from (1),

u'll get

 

S-S/(2n+1)²= 1/2n+1 - [2/(2n+1)³ +2/(2n+1)⁵-..........................+2/(2n+1)^(2n-1) ]

 - 1/(2n-1)(2n+1)⁽²ⁿ⁺¹⁾

 

Now, the highlighted part is a GP (take out 2 common first), with n-1 terms, so find its sum, and simplify the obtained expression. I'm not solving it completely here, so tell me if there's a problem, i'll complete the whole solution.

Thank u

 

Hey Ankita,

           Hello,

                Is the answer

                         16n^4+32^3+24n^2+4n+1  is correct??