JEE Main 2026 (Online) 2nd April Evening ShiftQuestion 1
The sum $\frac{1^{3}}{1}+\frac{1^{3}+2^{3}}{1+3}+\frac{1^{3}+2^{3}+3^{3}}{1+3+5}+\ldots$ up to $8$ terms, is :
Sequence and Series · Standard Sums (Σ formulas)
Solution
Formulas used. (1) $\sum_{k=1}^{n}k^3=\left[\frac{n(n+1)}{2}\right]^2$(2) $1+3+5+\cdots+(2n-1)=n^2$(3) $\sum_{k=1}^{n}k^2=\frac{n(n+1)(2n+1)}{6}$Step 1 - General term. The numerator is $1^3+2^3+\cdots+n^3=\left[\dfrac{n(n+1)}2\right]^2$ and the denominator is $1+3+5+\cdots+(2n-1)=n^2$: $$T_n=\frac{\dfrac{n^2(n+1)^2}{4}}{n^2}=\frac{(n+1)^2}{4}.$$ Step 2 - Sum of $8$ terms. $$S_8=\frac14\sum_{n=1}^{8}(n+1)^2=\frac14\left(2^2+3^2+\cdots+9^2\right).$$ Step 3 - Evaluate. $\sum_{k=1}^{9}k^2=\dfrac{9\cdot10\cdot19}{6}=285$, so removing $1^2$: $$S_8=\frac{285-1}{4}=\frac{284}{4}=71 .$$ Answer: (B)