JEE Main 2023 (Online) 1st February Morning ShiftQuestion 1
The sum to $10$ terms of the series $\frac{1}{1+1^2+1^4}+\frac{2}{1+2^2+2^4}+\frac{3}{1+3^2+3^4}+\ldots$ is :
Sequence and Series · Telescoping ($V_n$ Method)
Solution
Formulas used. (1) $n^4+n^2+1=(n^2+1)^2-n^2=(n^2-n+1)(n^2+n+1)$(2) Telescoping: $\sum_{n=1}^{N}(V_n-V_{n+1})=V_1-V_{N+1}$Step 1 - Partial fractions. The two factors differ by $2n$: $$\frac{n}{n^4+n^2+1}=\frac12\left[\frac1{n^2-n+1}-\frac1{n^2+n+1}\right].$$ Step 2 - Telescope. With $V_n=\dfrac1{n^2-n+1}$, $V_{n+1}=\dfrac1{n^2+n+1}$: $$\sum_{n=1}^{10}=\frac12\left(V_1-V_{11}\right)=\frac12\left(1-\frac1{111}\right)=\frac12\cdot\frac{110}{111}=\frac{55}{111}.$$ Answer: (B)