Question

The sum to \(n\) terms of the series \(\sum_{r=1}^n \frac{r}{1+r^2+r^4}\) is

• A. \(\frac{n^2+1}{2(n^2+n+1)}\)
• B. \(\frac{n^2+n}{2(n^2+n+1)}\)
• C. \(\frac{n^2+n}{2(n^2+n+1)}\)
• D. \(\frac{n^2+n}{2(n^3+n+1)}\)

Hint:

Use partial fractions: \(\frac{r}{1+r^2+r^4} = \frac{1}{2}\left[\frac{1}{r^2-r+1} - \frac{1}{r^2+r+1}\right]\)

Answer 1

The Correct Option is C

Step 1: Formula / Definition}
\[ T_r = \frac{r}{1+r^2+r^4} = \frac{r}{(r^2+1)^2 - r^2} = \frac{r}{(r^2-r+1)(r^2+r+1)} \]
Step 2: Calculation / Simplification}
\[ T_r = \frac{1}{2}\left[\frac{1}{r^2-r+1} - \frac{1}{r^2+r+1}\right] \]
\[ \sum_{r=1}^n T_r = \frac{1}{2}\left[\left(\frac{1}{1} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{7}\right) + \cdots + \left(\frac{1}{n^2-n+1} - \frac{1}{n^2+n+1}\right)\right] \]
\[ = \frac{1}{2}\left[1 - \frac{1}{n^2+n+1}\right] = \frac{1}{2}\left[\frac{n^2+n}{n^2+n+1}\right] \]
Step 3: Final Answer
\[ \frac{n^2+n}{2(n^2+n+1)} \]