JEE Mains · Maths · STD 11 - 8. sequence and series
The sum of the series \(\frac{1}{1-3 \cdot 1^2+1^4}+\) \(\frac{2}{1-3 \cdot 2^2+2^4}+\frac{3}{1-3 \cdot 3^2+3^4}+\ldots\). up to \(10\) terms is
- A \(\frac{45}{109}\)
- B \(-\frac{45}{109}\)
- C \(\frac{55}{109}\)
- D \(-\frac{55}{109}\)
Answer & Solution
Correct Answer
(D) \(-\frac{55}{109}\)
Step-by-step Solution
Detailed explanation
General term of the sequence, \(\mathrm{T}_{\mathrm{r}}=\frac{\mathrm{r}}{1-3 \mathrm{r}^2+\mathrm{r}^4}\) \(\mathrm{~T}_{\mathrm{r}}=\frac{\mathrm{r}}{\mathrm{r}^4-2 \mathrm{r}^2+1-\mathrm{r}^2}\)…
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