JEE Main 2026 · 2 Apr · Shift 2 · Mathematics Q5

MathematicsSingle correct+4 / -1Question 5 of 75

The sum $\dfrac{1^3}1+\dfrac{1^3+2^3}{1+3}+\dfrac{1^3+2^3+3^3}{1+3+5}+\cdots$ up to 8 terms, is :

Answer and worked solution

Official answer: B

General term. $T_n=\dfrac{\left(\frac{n(n+1)}2\right)^2}{n^2}=\dfrac{(n+1)^2}4$ .

Sum. $\displaystyle\sum_{n=1}^8\frac{(n+1)^2}4=\frac{2^2+\cdots+9^2}4=\frac{284}4=71$ .

(B)

Topics:Special series sums

JEE Main 2026 · 2 Apr · Shift 2, Mathematics question 5