Correct option is: (1) 7240
Given sum is
\(\mathrm{S}_{\mathrm{n}}=1+3+11+25+45+71+\ldots+\mathrm{T}_{\mathrm{n}}\)
First order differences are in A.P.
Thus, we can assume that
\(\mathrm{T}_{\mathrm{n}}=\mathrm{an}^{2}+\mathrm{bn}+\mathrm{c}\)
Solving \(\left\{\begin{array}{c}T_{1}=1=a+b+c \\ T_{2}=3=4 a+2 b+c \\ T_{3}=11=9 a+3 b+c\end{array}\right\},\)
we get \(\mathrm{a}=3, \mathrm{~b}=-7, \mathrm{c}=5\)
Hence, general term of given series is
\(\mathrm{T}_{\mathrm{n}}=3 \mathrm{n}^{2}-7 \mathrm{n}+5\)
Hence, required sum equals
\(\sum_\limits{n=1}^{n=20}\left(3 n^{2}-7 n+5\right)=3\left(\frac{20 \cdot 21 \cdot 41}{6}\right)-7\left(\frac{20 \cdot 21}{2}\right)+5(20)=7240\)
