Animated Solution for Mathematics - Sequence and Series: The sum of the series $\frac{1}{1-3\cdot 1^2+1^4}+\frac{2}{1-3\cdot 2^2+2^4}+\frac{3}{1-3\cdot 3^2+3^4}+\dots$ up to 10 terms is

The series ${\sum }_{r=1}^{10}\frac{r}{r^4-3r^2+1}$ appears intimidating at first glance. However, mathematics is rarely about brute force; it is about finding the hidden symmetry within the structure.

Let us begin by taming the denominator, $r^4-3r^2+1$. By treating $r^2$ as the primary variable, we can complete the square.

We split the term $-3r^2$ into $-2r^2-r^2$. This allows us to rewrite the expression as:

This transformation converts a complex quartic polynomial into a difference of two squares.

Using the algebraic identity $a^2-b^2=(a-b)(a+b)$, where $a=r^2-1$ and $b=r$, we factor the denominator:

Rearranging these terms for clarity, we obtain:

This step is the heart of the problem, as it reduces a degree-four polynomial into two manageable quadratic factors.

Next, we examine the numerator $r$ in relation to our new factors. Observe the difference between the two factors:

Because the difference is exactly twice our numerator, we can rewrite the general term $T_r$ as:

Splitting this expression yields the elegant form:

Now, we observe the series collapse through the telescoping effect. Let us evaluate the first few terms:

For $r=1$: $T_1=\frac{1}{2}[\frac{1}{-1}-\frac{1}{1}]$

For $r=2$: $T_2=\frac{1}{2}[\frac{1}{1}-\frac{1}{5}]$

For $r=3$: $T_3=\frac{1}{2}[\frac{1}{5}-\frac{1}{11}]$

The terms cancel sequentially: the $-\frac{1}{1}$ from $T_1$ cancels with the $+\frac{1}{1}$ from $T_2$, and the $-\frac{1}{5}$ from $T_2$ cancels with the $+\frac{1}{5}$ from $T_3$. Only the first part of $T_1$ and the second part of $T_{10}$ remain.

The survivors of this collapse are $\frac{1}{2}(-1)$ and $-\frac{1}{2}(\frac{1}{10^2+10-1})$. Calculating the final term:

Summing these values, we get:

The final result of the series is $-\frac{55}{109}$.