JEE MAIN - Mathematics (2024 - 31st January Morning Shift - No. 10)

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
$$\frac{45}{109}$$
$$-\frac{55}{109}$$
$$\frac{55}{109}$$
$$-\frac{45}{109}$$

Explanation

General term of the sequence,

$$\begin{aligned} & \mathrm{T}_{\mathrm{r}}=\frac{\mathrm{r}}{1-3 \mathrm{r}^2+\mathrm{r}^4} \\ & \mathrm{~T}_{\mathrm{r}}=\frac{\mathrm{r}}{\mathrm{r}^4-2 \mathrm{r}^2+1-\mathrm{r}^2} \\ & \mathrm{~T}_{\mathrm{r}}=\frac{\mathrm{r}}{\left(\mathrm{r}^2-1\right)^2-\mathrm{r}^2} \\ & \mathrm{~T}_{\mathrm{r}}=\frac{\mathrm{r}}{\left(\mathrm{r}^2-\mathrm{r}-1\right)\left(\mathrm{r}^2+\mathrm{r}-1\right)} \\ & \mathrm{T}_{\mathrm{r}}=\frac{\frac{1}{2}\left[\left(\mathrm{r}^2+\mathrm{r}-1\right)-\left(\mathrm{r}^2-\mathrm{r}-1\right)\right]}{\left(\mathrm{r}^2-\mathrm{r}-1\right)\left(\mathrm{r}^2+\mathrm{r}-1\right)} \\ & =\frac{1}{2}\left[\frac{1}{\mathrm{r}^2-\mathrm{r}-1}-\frac{1}{\mathrm{r}^2+\mathrm{r}-1}\right] \end{aligned}$$

Sum of 10 terms,

$$\sum_\limits{\mathrm{r}=1}^{10} \mathrm{~T}_{\mathrm{r}}=\frac{1}{2}\left[\frac{1}{-1}-\frac{1}{109}\right]=\frac{-55}{109}$$

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