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Mathematics Question on Sequence and series

The sum of the series 11−3⋅12+14+21−3⋅22+24+31−3⋅32+34+… up to 10 terms\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 \text{ up to 10 terms} is

A

45109\frac{45}{109}

B

−45109-\frac{45}{109}

C

55109\frac{55}{109}

D

−55109-\frac{55}{109}

Answer

−55109-\frac{55}{109}

Explanation

Solution

General term of the sequence,

Tr=r1−3r2+r4T_r = \frac{r}{1 - 3r^2 + r^4}

Tr=rr4−3r2+1−r2T_r = \frac{r}{r^4 - 3r^2 + 1 - r^2}

Tr=r(r2−1)2−r2T_r = \frac{r}{(r^2 - 1)^2 - r^2}

Tr=r(r2−r−1)(r2+r−1)T_r = \frac{r}{(r^2 - r - 1)(r^2 + r - 1)}

Tr=12[1(r2−r−1)−1(r2+r−1)]T_r = \frac{1}{2} \left[ \frac{1}{(r^2 - r - 1)} - \frac{1}{(r^2 + r - 1)} \right]

Sum of 10 terms,

∑r=110Tr=12[1−1−1109]=55109\sum_{r=1}^{10} T_r = \frac{1}{2} \left[ \frac{1}{-1} - \frac{1}{109} \right] = \frac{55}{109}