The sum (1^3)/(1) + (1^3 + 2^3)/(1 + 3) + (1^3 + 2^3 + 3^3)/(1 + 3 + 5) + up to 8 terms, is:

Correct answer: B. 71 The n -th term of the given series is: T_n = (1^3 + 2^3 + + n^3)/(1 + 3 + 5 + + (2n - 1)) Using the formulas for the sum of cubes of first n natural numbers and the sum of first n odd numbers: T_n = (( (n(n+1))/(2) )^2)/(n^2) = (n^2(n+1)^2)/(4n^2) = ((n+1)^2)/(4) The sum of the series up to 8 terms is: S_8 = ∑_ n=1 ^ 8 T_n = ∑_ n=1 ^ 8 ((n+1)^2)/(4) S_8 = (1)/(4) ( 2^2 + 3^2 + + 9^2 ) S_8 = (1)/(4) ( ∑_ k=1 ^ 9 k^2 - 1^2 ) Using the formula ∑_ k=1 ^ n k^2 = (n(n+1)(2n+1))/(6) : S_8 = (1)/(4) ( (9 × 10 × 19)/(6) - 1 ) S_8 = (1)/(4) (285 - 1) = (284)/(4) = 71 Answer: 71

JEE Main 20262 AprilEvening Shift

Exam: JEE Main
Year: 2026
Date: 2 April
Shift: Evening Shift
Subject: Mathematics
Chapter: Sequences and Series

JEE Main 2026 · 2 April · Evening Shift · Mathematics · Sequences and Series

The sum (1^3)/(1) + (1^3 + 2^3)/(1 + 3) + (1^3 + 2^3 + 3^3)/(1 + 3...

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:

Options

A $70$
B $71$
C $72$
D $73$

Correct answer

B. $71$

Step-by-step solution

The

n

-th term of the given series is:

T_n = \dfrac{1^3 + 2^3 + \cdots + n^3}{1 + 3 + 5 + \cdots + (2n - 1)}

Using the formulas for the sum of cubes of first

n

natural numbers and the sum of first

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}

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}

S_8 = \dfrac{1}{4} \left( 2^2 + 3^2 + \cdots + 9^2 \right)

S_8 = \dfrac{1}{4} \left( \sum_{k=1}^{9} k^2 - 1^2 \right)

Using the formula

\sum_{k=1}^{n} k^2 = \dfrac{n(n+1)(2n+1)}{6}

S_8 = \dfrac{1}{4} \left( \dfrac{9 \times 10 \times 19}{6} - 1 \right)

S_8 = \dfrac{1}{4} (285 - 1) = \dfrac{284}{4} = 71

Answer:

71

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