Induction Time

S(k) as the sum of infinite series first term k and common ratio 1/k+1

                                     S(k) =  k/(1-1/k+1)  =  K+1  
                                     S(1) square = 2  square  
                                      S(2)        ''       = 3      ''
                                     S(3)        ''         = 4      ''

                                       S(2n-1)   ''     =  (2n)   ''

                                     summing up results  =  2 square +  3  square + 4 square  +
                                                                                               +++            +(2n) square

do one thing and 1 squre and subtract one square then it will turn into suare summation (2n)(2n+1)(4n+1)/6 -1 = 1/3 ( (n)(2n+1)(4n+1) - 3)

try every problem through simple aaproach :) :)

$S_k=\cfrac{k}{1-\frac{1}{k+1}}=k+1$ .

Thus, we have:

$S_1^2+S_2^2+\ldots+S_{2n-1}^2=2^2+3^2+\ldots+(2n)^2$

Since $1^2+2^2+\ldots+k^2=\frac{n(n+1)(2n+1)}{6}$

We have the $2^2+3^2+\ldots+(2n)^2=\frac{2n(2n+1)(4n+1)}{6}-1$

Simplifying, we have our desired sum of $\boxed{\frac{n(2n+1)(4n+1)-3}{3}}$