Mathematics - Sequence & Series Question with Solution | TestHub

Given sum is
$S_n=1+3+11+25+45+71+\dots +T_n$
First order differences are in A.P.
Thus, we can assume that
${\mathrm{T}}_{\mathrm{n}}={\mathrm{an}}^2+\mathrm{bn}+\mathrm{c}$
Solving $\left\{\begin{array}{c}{T}_1=1=a+b+c\\ T_2=3=4a+2b+c\\ T_3=11=9a+3b+c\end{array}\right\}$,
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
${\sum }_{\mathrm{n}=1}^{\mathrm{n}=20}\left(3{\mathrm{n}}^2-7\mathrm{n}+5\right)=3\left(\frac{20\cdot 21\cdot 41}{6}\right)-7\left(\frac{20\cdot 21}{2}\right)+5(20)=7240$