JEE Main 2025 (Online) 4th April Evening ShiftQuestion 1
If the sum of the first $20$ terms of the series $\frac{4\cdot1}{4+3\cdot1^{2}+1^{4}}+\frac{4\cdot2}{4+3\cdot2^{2}+2^{4}}+\frac{4\cdot3}{4+3\cdot3^{2}+3^{4}}+\frac{4\cdot4}{4+3\cdot4^{2}+4^{4}}+\ldots$ is $\frac{m}{n}$, where $m$ and $n$ are coprime, then $m+n$ is equal to :
Sequence and Series · Telescoping ($V_n$ Method)
Solution
Formulas used. (1) $n^4+3n^2+4=(n^2+2)^2-n^2=(n^2-n+2)(n^2+n+2)$(2) Telescoping: $\sum_{n=1}^{N}(V_n-V_{n+1})=V_1-V_{N+1}$Step 1 - Factorise. $$4+3n^2+n^4=(n^2+2)^2-n^2=(n^2-n+2)(n^2+n+2).$$ Step 2 - Partial fractions. The two factors differ by $2n$, so $$\frac{4n}{(n^2-n+2)(n^2+n+2)}=2\left[\frac1{n^2-n+2}-\frac1{n^2+n+2}\right].$$ Step 3 - Telescope. Put $V_n=\dfrac1{n^2-n+2}$. Then $V_{n+1}=\dfrac1{(n+1)^2-(n+1)+2}=\dfrac1{n^2+n+2}$, so the term is $2(V_n-V_{n+1})$. $$\sum_{n=1}^{20}=2\left(V_1-V_{21}\right)=2\left(\frac12-\frac1{422}\right)=1-\frac1{211}=\frac{210}{211}.$$ Step 4. $211$ is prime, so $m=210,\ n=211$ and $$m+n=421 .$$ Answer: (B)