JEE Main 2026 (Online) 5th April Morning ShiftQuestion 1
$\sum_{n=1}^{10}\left(\frac{528}{n(n+1)(n+2)}\right)$ is equal to :
Sequence and Series · Telescoping ($V_n$ Method)
Solution
Formulas used. (1) Partial fractions: $\frac{1}{n(n+1)(n+2)}=\frac12\left[\frac1{n(n+1)}-\frac1{(n+1)(n+2)}\right]$(2) Telescoping: $\sum_{n=1}^{N}(V_n-V_{n+1})=V_1-V_{N+1}$Step 1 - Split the general term. $$\frac1{n(n+1)(n+2)}=\frac12\left[\frac1{n(n+1)}-\frac1{(n+1)(n+2)}\right].$$ (Check: the bracket equals $\dfrac{(n+2)-n}{n(n+1)(n+2)}=\dfrac{2}{n(n+1)(n+2)}$.) Step 2 - Telescope. Put $V_n=\dfrac1{n(n+1)}$; the term is $\tfrac12(V_n-V_{n+1})$. $$\sum_{n=1}^{10}\frac1{n(n+1)(n+2)}=\frac12\left(V_1-V_{11}\right)=\frac12\left(\frac12-\frac1{132}\right)=\frac12\cdot\frac{65}{132}=\frac{65}{264}.$$ Step 3 - Multiply by $528$. $$528\cdot\frac{65}{264}=2\cdot65=130 .$$ Answer: (B)