Question

Find sum up to 8 terms of the series
\[ \frac{1^3}{1}+\frac{1^3+2^3}{1+3}+\frac{1^3+2^3+3^3}{1+3+5}+\cdots \]

• A. 84
• B. 71
• C. 61
• D. 100

Hint:

In such series, first identify the general term. Here, use the two important formulas: sum of cubes \(=\left(\frac{n(n+1)}{2}\right)^2\) and sum of first \(n\) odd numbers \(=n^2\). This simplifies the term immediately.

Answer 1

The Correct Option is B

Step 1: Express the general term of the series.
The \( n \)th term of the given series is: \[ T_n = \frac{1^3 + 2^3 + 3^3 + \cdots + n^3} {1 + 3 + 5 + \cdots + (2n - 1)}. \]
Step 2: Apply standard summation formulas.
We use the known results: \[ 1^3 + 2^3 + 3^3 + \cdots + n^3 = \left( \frac{n(n+1)}{2} \right)^2, \] and the sum of the first \( n \) odd numbers: \[ 1 + 3 + 5 + \cdots + (2n - 1) = n^2. \] Substituting these expressions into \( T_n \): \[ T_n = \frac{ \left( \frac{n(n+1)}{2} \right)^2 }{n^2}. \] Simplifying: \[ T_n = \frac{n^2 (n+1)^2}{4n^2} = \frac{(n+1)^2}{4}. \]
Step 3: Compute the sum of the first 8 terms.
Thus, the required sum is: \[ S_8 = \sum_{n=1}^{8} \frac{(n+1)^2}{4}. \] \[ S_8 = \frac{1}{4} \sum_{n=1}^{8} (n+1)^2. \] Let \( k = n + 1 \). Then as \( n \) varies from 1 to 8, \( k \) varies from 2 to 9. Therefore: \[ S_8 = \frac{1}{4} \sum_{k=2}^{9} k^2. \] Using the formula for the sum of squares: \[ \sum_{k=1}^{9} k^2 = \frac{9 \times 10 \times 19}{6} = 285. \] Hence, \[ \sum_{k=2}^{9} k^2 = 285 - 1 = 284. \] So, \[ S_8 = \frac{284}{4} = 71. \]
Step 4: Conclusion.
Therefore, the sum of the first 8 terms of the series is:
Final Answer: \( 71 \).