JEE Main 2026 — Mathematics Algebra

The $n$n-th term of the given series is:

$$T_n = \dfrac{1^3 + 2^3 + \cdots + n^3}{1 + 3 + 5 + \cdots + (2n - 1)}$$Tn​=1+3+5+⋯+(2n−1)13+23+⋯+n3​

Using the formulas for the sum of cubes of first $n$n natural numbers and the sum of first $n$n odd numbers:

$$T_n = \dfrac{\left( \dfrac{n(n+1)}{2} \right)^2}{n^2} = \dfrac{n^2(n+1)^2}{4n^2} = \dfrac{(n+1)^2}{4}$$Tn​=n2(2n(n+1)​)2​=4n2n2(n+1)2​=4(n+1)2​

The sum of the series up to 8 terms is:

$$S_8 = \sum_{n=1}^{8} T_n = \sum_{n=1}^{8} \dfrac{(n+1)^2}{4}$$S8​=n=1∑8​Tn​=n=1∑8​4(n+1)2​

$$S_8 = \dfrac{1}{4} \left( 2^2 + 3^2 + \cdots + 9^2 \right)$$S8​=41​(22+32+⋯+92)

$$S_8 = \dfrac{1}{4} \left( \sum_{k=1}^{9} k^2 - 1^2 \right)$$S8​=41​(k=1∑9​k2−12)

Using the formula $\displaystyle \sum_{k=1}^{n} k^2 = \dfrac{n(n+1)(2n+1)}{6}$k=1∑n​k2=6n(n+1)(2n+1)​:

$$S_8 = \dfrac{1}{4} \left( \dfrac{9 \times 10 \times 19}{6} - 1 \right)$$S8​=41​(69×10×19​−1)

$$S_8 = \dfrac{1}{4} (285 - 1) = \dfrac{284}{4} = 71$$S8​=41​(285−1)=4284​=71

Answer: $71$71