The sum of the series $$\frac{1}{1 - 3 \cdot 1^2 + 1^4} + \frac{2}{1 - 3 \cdot 2^2 + 2^4} + \frac{3}{1 - 3 \cdot 3^2 + 3^4} + \ldots$$ up to 10 terms is

The general term of the series can be written as:

$$T_n = \frac{n}{1 - 3n^2 + n^4}$$

First, let's factor the denominator by completing the square:

$$n^4 - 3n^2 + 1 = n^4 - 2n^2 + 1 - n^2$$ $$= (n^2 - 1)^2 - n^2$$

Using the identity $$a^2 - b^2 = (a-b)(a+b)$$:

$$= (n^2 - 1 - n)(n^2 - 1 + n) = (n^2 - n - 1)(n^2 + n - 1)$$

Now, rewrite $$T_n$$ using these factors. Notice that the difference between the two terms in the denominator is $$(n^2 + n - 1) - (n^2 - n - 1) = 2n$$, which is twice our numerator. We can use this to split the term into partial fractions:

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

$$T_n = \frac{1}{2} \left[ \frac{2n}{(n^2 - n - 1)(n^2 + n - 1)} \right]$$

$$T_n = \frac{1}{2} \left[ \frac{(n^2 + n - 1) - (n^2 - n - 1)}{(n^2 - n - 1)(n^2 + n - 1)} \right]$$

$$T_n = \frac{1}{2} \left[ \frac{1}{n^2 - n - 1} - \frac{1}{n^2 + n - 1} \right]$$

This forms a telescoping series. Notice that if we let $$V_n = \frac{1}{n^2 - n - 1}$$, then the next term $$V_{n+1}$$ is:

$$V_{n+1} = \frac{1}{(n+1)^2 - (n+1) - 1} = \frac{1}{n^2 + 2n + 1 - n - 1 - 1} = \frac{1}{n^2 + n - 1}$$

So, $$T_n = \frac{1}{2}(V_n - V_{n+1})$$.

Summing up to 10 terms:

$$S_{10} = \sum_{n=1}^{10} T_n = \frac{1}{2} [ (V_1 - V_2) + (V_2 - V_3) + \dots + (V_{10} - V_{11}) ]$$

Most terms cancel out, leaving just the first and last terms:

$$S_{10} = \frac{1}{2} (V_1 - V_{11})$$

Substitute $$n=1$$ and $$n=11$$ to find $$V_1$$ and $$V_{11}$$:

• $$V_1 = \frac{1}{1^2 - 1 - 1} = -1$$
• $$V_{11} = \frac{1}{11^2 - 11 - 1} = \frac{1}{121 - 12} = \frac{1}{109}$$

Finally, calculate the sum:

$$S_{10} = \frac{1}{2} \left( -1 - \frac{1}{109} \right) = \frac{1}{2} \left( \frac{-109 - 1}{109} \right) = \frac{1}{2} \left( \frac{-110}{109} \right) = -\frac{55}{109}$$