Question

The sum \( \frac{1^3}{1} + \frac{2^3}{1+3} + \frac{3^3}{1+3+5} + \cdots \) up to 8 terms is:

• A. 70
• B. 71
• C. 72
• D. 73

Answer 1

The Correct Option is B

We are given a series where the general term is: \[ T_n = \frac{1^3 + 2^3 + \cdots + n^3}{1 + 3 + 5 + \cdots + (2n - 1)}. \] The sum of cubes of the first \( n \) natural numbers is given by the formula: \[ 1^3 + 2^3 + \cdots + n^3 = \left( \frac{n(n + 1)}{2} \right)^2. \] The sum of the first \( n \) odd numbers is: \[ 1 + 3 + 5 + \cdots + (2n - 1) = n^2. \] Thus, the general term \( T_n \) simplifies to: \[ T_n = \frac{\left( \frac{n(n + 1)}{2} \right)^2}{n^2} = \frac{(n(n + 1))^2}{4n^2} = \frac{(n + 1)^2}{4}. \] Now, we can calculate the sum of the first 8 terms: \[ S = \sum_{n=1}^{8} T_n = \sum_{n=1}^{8} \frac{(n + 1)^2}{4}. \] Evaluating the sum: \[ S = \frac{(2)^2}{4} + \frac{(3)^2}{4} + \frac{(4)^2}{4} + \cdots + \frac{(9)^2}{4} = \frac{4 + 9 + 16 + 25 + 36 + 49 + 64 + 81}{4}. \] \[ S = \frac{284}{4} = 71. \] Thus, the sum of the series up to 8 terms is 71.
Final Answer: 71