Sequences And Series Question 16
Question 16 - 2024 (31 Jan Shift 1)
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
(1) $\frac{45}{109}$
(2) $-\frac{45}{109}$
(3) $\frac{55}{109}$
(4) $-\frac{55}{109}$
Show Answer
Answer (4)
Solution
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,
$\sum_{\mathrm{r}=1}^{10} \mathrm{~T}_{\mathrm{r}}=\frac{1}{2}\left[\frac{1}{-1}-\frac{1}{109}\right]=\frac{-55}{109}$
