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

Step 1: Understanding the Concept:
We first need to find the general \(n\)-th term \(T_n\) of the given series.
The numerator of the \(n\)-th term is the sum of the cubes of the first \(n\) natural numbers.
The denominator is the sum of the first \(n\) odd numbers.
Step 2: Key Formula or Approach:
The formula for the numerator sum is \(1^3 + 2^3 + \dots + n^3 = \left( \frac{n(n+1)}{2} \right)^2 = \frac{n^2(n+1)^2}{4}\).
The formula for the denominator sum is \(1 + 3 + 5 + \dots + (2n-1) = n^2\).
Thus, the general term is given by \(T_n = \frac{n^2(n+1)^2 / 4}{n^2} = \frac{(n+1)^2}{4}\).
Step 3: Detailed Explanation:
We need the sum up to 8 terms, which is \(S_8 = \sum_{n=1}^8 T_n\).
\[ S_8 = \sum_{n=1}^8 \frac{(n+1)^2}{4} = \frac{1}{4} \sum_{n=1}^8 (n+1)^2 \] To evaluate this efficiently, let \(k = n + 1\).
As \(n\) goes from \(1\) to \(8\), \(k\) goes from \(2\) to \(9\).
\[ S_8 = \frac{1}{4} \sum_{k=2}^9 k^2 \] We know the sum of squares formula is \(\sum_{k=1}^m k^2 = \frac{m(m+1)(2m+1)}{6}\).
For \(m = 9\), we calculate the total sum.
\[ \sum_{k=1}^9 k^2 = \frac{9 \times 10 \times 19}{6} = 15 \times 19 = 285 \] Since our sum starts from \(k = 2\), we subtract the \(k = 1\) term which is \(1^2 = 1\).
\[ \sum_{k=2}^9 k^2 = 285 - 1 = 284 \] Now, calculate the final sum \(S_8\).
\[ S_8 = \frac{1}{4} \times 284 = 71 \] Step 4: Final Answer:
The sum up to 8 terms is \(71\).