JEE Main 2025 — 3 April, Morning Shift — Question 26

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 $a=3, b=-7, c=5$

Hence, general term of given series is

$T_{n}=3 n^{2}-7 n+5$

Hence, required sum equals

$\sum_{n=1}^{n=20}\left(3 n^{2}-7 n+5\right)=3\left(\frac{20 \cdot 21 \cdot 41}{6}\right)-7\left(\frac{20 \cdot 21}{2}\right)+5(20)=7240$