The question is below?

Here ,

#S=1^2/1+(1+2)^2/(1+3)+(1+2+3)^2/(1+3+5)+(1+2+3+4)^2/(1+3+5+7)+...#

Let , # r^(th)term =t_r ,then#

#t_r=(1+2+3+...+r)^2/(1+3+5+...+(2r-1))=(sumr)^2/(sum(2r-1))#

#:.t_r=[r/2(r+1)]^2/(2sumr-sum1)=[r^2/4(r+1)^2]/[2(r/2(r+1))-r)=[r^2/4(r+1)^2]/(r(r+1-1)#

#:.t_r=[r^2/4(r+1)^2]/r^2=1/4(r^2+2r+1)#

So,

#S_n=sum_(r=1)^n1/4[r^2+2r+1]=1/4{sum_(r=1)^nr^2+2sum_(r=1)^nr+sum_(r=1)^n1}#

#:.S_n=1/4{n/6(n+1)(2n+1)+2*n/2(n+1)+n}#

#:.S_n=1/4xxn/6{(n+1)(2n+1)+6(n+1)+6}#

#=>S_n=n/24{2n^2+n+2n+1+6n+6+6}#

#=>S_n=n/24{2n^2+9n+13}#

If you look at #n#th term, it is the square of the sum of the first #n# integers divided by the sum of the first #n# odd integers.

Recall that the sum of the first #n# integers is #1/2n(n+1)#. Squaring this gives #1/4n^2(n+1)^2#.

Now, the sum of the first #n# odd integers is #n^2#.

Thus, we have each term being #1/4(n+1)^2#. Sum this:

#sum_(n=1)^N1/4(n+1)^2#

#=1/4sum_(n=1)^N(n+1)^2#

#=1/4(sum_(n=1)^(N+1)n^2-1)#

Recall the formula for the sum of the first #n# perfect squares:

#=1/4((2(N+1)^3+3(N+1)^2+(N+1))/6-1)#

#=(2N^3+9N^2+13N)/24#