Sequences & Series
Telescoping Series — Sum of 10 Terms
nta_pyq_2024_jan
Grade 11
Question:
The sum of the series $\dfrac{1}{1-3\cdot1^2+1^4}+\dfrac{2}{1-3\cdot2^2+2^4}+\dfrac{3}{1-3\cdot3^2+3^4}+\cdots$ up to 10 terms is
$\dfrac{45}{109}$
$-\dfrac{45}{109}$
$\dfrac{55}{109}$
$-\dfrac{55}{109}$
Step-by-Step Solution
Key Concept: General term: $T_r=\frac{r}{1-3r^2+r^4}=\frac{r}{(r^2-r-1)(r^2+r-1)}=\frac{1}{2}\left(\frac{1}{r^2-r-1}-\frac{1}{r^2+r-1}\right)$. This telescopes.
Telescoping gives $\frac{1}{2}(-1-\frac{1}{109})=-\frac{55}{109}$.
Correct Answer: 4