The sum 1+3+11+25+45+71+. . upto 20 terms, is equal to

Given sum is
$S_n=1+3+11+25+45+71+\ldots+T_n$
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\{$$\begin{array}{c}T_1=1=a+b+c \\ T_2=3=4 a+2 b+c \\ T_3=11=9 a+3 b+c\end{array}$$,
we get \mathrm{a}=3, \mathrm{~b}=-7, \mathrm{c}=5$<br/>Hence, general term of given series is<br/>$\mathrm{T}_{\mathrm{n}}=3\mathrm{n}^2-7\mathrm{n}+$5$
Hence, required sum equals
\sum_{\mathrm{n}=1}^{\mathrm{n}=20}\left(3 \mathrm{n}^2-7$\mathrm{n}$+5\right)=3 \left(\frac{20 \cdot 21 \cdot 41}{6}\right)$-7 \left(\frac{20 \cdot 21}{2}\right)$+5(20)=7240