JEE 2023 — Mathematics PYQ

$S_n=5+11+19+29+41+\cdots +T_n$

$S_n=5+11+19+29+\cdots +T_{n-1}+T_n$

$0=5+6+8+10+12+\cdots -T_n$

$T_n=\frac{n-1}{2}[2\times 6+(n-2)\times 2]$

$T_n=n^2+3n+1$

$\Rightarrow S_n=\sum T_n=\sum (n^2+3n+1)$

$S_{20}=3520$