Given sum is
Sn=1+3+11+25+45+71+…+Tn
First order differences are in A.P.
Thus, we can assume that
Tn=an2+bn+c
Solving ⎩⎨⎧T1=1=a+b+cT2=3=4a+2b+cT3=11=9a+3b+c⎭⎬⎫,
we get a=3, b=−7,c=5
Hence, general term of given series is
Tn=3n2−7n+5
Hence, required sum equals
n=1∑n=20(3n2−7n+5)=3(620⋅21⋅41)−7(220⋅21)+5(20)=7240