If first terms of a decreasing infinite GP is 1 and sum is S, then sum of squares of its terms is??

sum of infinite term in  G.P is a/(1-r). (where a is 1st term, and r is the ratio)

so here in this problem, 1/(1-r)=S    (Since a=1)

solving it u will get; r =1 - 1/s

since terms of a G.P are given by a, ar ,ar^2,ar^3..........and so on

so here u will get the following series...

i.e 1 + (1- 1/s) + (1-1/s)^2 + (1-1/s)^3+..........

on squaring each term u will get,

1^2 + (1- 1/s)^2 + (1 - 1/s)^4 +(1-1/s)^6.........[r =(1-1/s)^2] in this case                                     (1)

so the required sum of this series  is 

s'=1 /(1 -r) substitute value of r from (1)

finally u will get s' =s^2/(2s-1) ans..   .hope my answer will satisfy u.

let the GP is a + ar + ar² + ar³ ...............infinity

sum of infinite GP S = a/1-r      

                         S = 1/1-r                     (a=1)

                        r = S-1/S       ..........1

now gp formed after squaring is a² + a²r² + a²r⁴  ...............infinity

its common ratio is r²

 now sum is S¹ = a/1-r² =1/(1-r)(1+r)

   from eq 1 putting value of r

          S¹ = S²/(2S-1)