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: Write 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: Use standard formulas for numerator and denominator.
We know that:
\[ 1^3+2^3+3^3+\cdots+n^3=\left(\frac{n(n+1)}{2}\right)^2 \] Also, the sum of first \(n\) odd numbers is:
\[ 1+3+5+\cdots+(2n-1)=n^2 \] So,
\[ T_n=\frac{\left(\frac{n(n+1)}{2}\right)^2}{n^2} \] \[ T_n=\frac{n^2(n+1)^2}{4n^2} \] \[ T_n=\frac{(n+1)^2}{4} \]
Step 3: Find the sum of first 8 terms.
Hence, 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 \(n\) goes from 1 to 8, \(k\) goes from 2 to 9. So:
\[ S_8=\frac{1}{4}\sum_{k=2}^{9}k^2 \] Now,
\[ \sum_{k=1}^{9}k^2=\frac{9 \cdot 10 \cdot 19}{6}=285 \] Therefore,
\[ \sum_{k=2}^{9}k^2=285-1=284 \] So,
\[ S_8=\frac{284}{4}=71 \]
Step 4: Conclusion.
Therefore, the sum up to 8 terms of the series is \(71\).
Final Answer: \(71\)