Understanding the Series Terms

The series given is:

$ \frac{1^3}{1} + \frac{1^3 + 2^3}{1 + 3} + \frac{1^3 + 2^3 + 3^3}{1 + 3 + 5} + \dots $

We need to determine the formula for the n-th term.

• The numerator of the n-th term is the sum of the cubes of the first n natural numbers: $ \sum_{i=1}^{n} i^3 = \left( \frac{n(n+1)}{2} \right)^2 $.
• The denominator of the n-th term is the sum of the first n odd natural numbers: $ \sum_{i=1}^{n} (2i-1) = n^2 $.

So, the n-th term ($T_n$) of the series is:

$ T_n = \frac{\text{Sum of first n cubes}}{\text{Sum of first n odd numbers}} = \frac{\left( \frac{n(n+1)}{2} \right)^2}{n^2} $

Simplifying this expression:

$ T_n = \frac{\frac{n^2(n+1)^2}{4}}{n^2} = \frac{(n+1)^2}{4} $

Calculating the Sum of 8 Terms

The question asks for the sum of the first 8 terms of this series. Using the formula derived for $T_n$:

$ S_8 = \sum_{n=1}^{8} T_n = \sum_{n=1}^{8} \frac{(n+1)^2}{4} $

We can factor out the constant $\frac{1}{4}$:

$ S_8 = \frac{1}{4} \sum_{n=1}^{8} (n+1)^2 $

To evaluate the sum $ \sum_{n=1}^{8} (n+1)^2 $, 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 $

The sum $ \sum_{k=2}^{9} k^2 $ can be calculated as $ \left( \sum_{k=1}^{9} k^2 \right) - 1^2 $.

Using the standard formula for the sum of the first N squares, $ \sum_{k=1}^{N} k^2 = \frac{N(N+1)(2N+1)}{6} $:

$ \sum_{k=1}^{9} k^2 = \frac{9(9+1)(2 \times 9 + 1)}{6} = \frac{9 \times 10 \times 19}{6} = \frac{1710}{6} = 285 $

Now, calculate $ \sum_{k=2}^{9} k^2 $:

$ \sum_{k=2}^{9} k^2 = 285 - 1^2 = 285 - 1 = 284 $

Finally, substitute this value back into the expression for $S_8$:

$ S_8 = \frac{1}{4} \times 284 = 71 $

Conclusion

The sum of the series up to 8 terms is 71.