JEE Main · 2025 · 04 Apr 2025 · Shift 1 · Sequences and Series

The $n$th term is the $n$th odd number $2n-1$, squared when $n$ is odd and taken as it is when $n$ is even. Over $40$ terms the odd positions contribute ${\sum }_{r=1}^{20}(4r-3{)}^2$ and the even positions contribute ${\sum }_{r=1}^{20}(4r-1)$, so the total is ${\sum }_{r=1}^{20}[(4r-3{)}^2+(4r-1)]={\sum }_{r=1}^{20}(16r^2-20r+8)$. Using ${\sum }_{r=1}^{20}{r}^2=2870$ and ${\sum }_{r=1}^{20}r=210$, this equals $16(2870)-20(210)+8(20)=45920-4200+160=41880$.