Verified Solution • 2025–26 Session moderate

QJEEM24Ma85ed: JEE Main 2024 Mathematics - Class 11 JEEM24Ma85ed Solution: The sum of the series up to 10 -terms is

The sum of the series 11−3⋅12+14+21−3⋅22+24+31−3⋅32+34+…\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

Multiple Choice Options

  • A 45109\frac{45}{109}
  • B −55109-\frac{55}{109}
  • C 55109\frac{55}{109}
  • D −45109-\frac{45}{109}

Step-by-Step Verified Solution

General term of the sequence,

amp;Tr=r1−3r2+r4amp; Tr=rr4−2r2+1−r2amp; Tr=r(r2−1)2−r2amp; Tr=r(r2−r−1)(r2+r−1)amp;Tr=12[(r2+r−1)−(r2−r−1)](r2−r−1)(r2+r−1)amp;=12[1r2−r−1−1r2+r−1]\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,

∑r=110 Tr=12[1−1−1109]=−55109

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