Question

Sum of \( \frac{1^3}{1} + \frac{1^3 + 2^3}{1 + 3} + \frac{1^3 + 2^3 + 3^3}{1 + 3 + 5} + \dots \) up to 8 terms is:

• A. 71
• B. 70
• C. 73
• D. 72

Hint:

The sum of the first \( n \) odd numbers is always \( n^2 \). This often simplifies series problems where the denominator looks complex.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
We need to find the general term (\( T_n \)) of the series. Each term is a ratio where the numerator is the sum of cubes of the first \( n \) natural numbers and the denominator is the sum of the first \( n \) odd numbers.
Step 2: Key Formula or Approach:
\[ \text{Numerator: } \sum n^3 = \left( \frac{n(n+1)}{2} \right)^2 \] \[ \text{Denominator: } 1 + 3 + 5 + \dots + (2n-1) = n^2 \]
Step 3: Detailed Explanation:
The \( n \)-th term \( T_n \) is: \[ T_n = \frac{\frac{n^2(n+1)^2}{4}}{n^2} = \frac{(n+1)^2}{4} = \frac{1}{4}(n^2 + 2n + 1) \] The sum of 8 terms is \( S_8 = \sum_{n=1}^{8} T_n \): \[ S_8 = \frac{1}{4} \left[ \sum_{n=1}^{8} n^2 + 2\sum_{n=1}^{8} n + \sum_{n=1}^{8} 1 \right] \] \[ \sum n^2 = \frac{8(9)(17)}{6} = 204 \] \[ 2\sum n = 2 \times \frac{8(9)}{2} = 72 \] \[ \sum 1 = 8 \] \[ S_8 = \frac{1}{4} [204 + 72 + 8] = \frac{1}{4} [284] = 71 \text{ (Wait, re-calculating: } 204+72+8 = 284; 284/4 = 71) \] *(Note: Recalculating based on term values: \( T_1=1, T_2=2.25, T_3=4, T_4=6.25, T_5=9, T_6=12.25, T_7=16, T_8=20.25 \). Sum: \( 1+4+9+16 = 30 \); \( 2.25+6.25+12.25+20.25 = 41 \). Total = 71).*
Step 4: Final Answer:
The sum is 71. (Correcting option mapping: if 71 is (a), then (a) is the answer).