JEE Main 2026 — 5 April, Morning Shift — Question 28

Let $\sum_{\mathrm{n}=1}^{10} \frac{528}{\mathrm{n}(\mathrm{n}+1)(\mathrm{n}+2)}=\lambda$ Let $\mathrm{T}_{\mathrm{n}}=\frac{1}{\mathrm{n}(\mathrm{n}+1)(\mathrm{n}+2)}$ $\mathrm{T}_{\mathrm{n}}=\frac{1}{2}\left\{\frac{(\mathrm{n}+2)-\mathrm{n}}{\mathrm{n}(\mathrm{n}+1)(\mathrm{n}+2)}\right\}$ $\mathrm{T}_{\mathrm{n}}=\frac{1}{2}\left\{\frac{1}{\mathrm{n}(\mathrm{n}+1)}-\frac{1}{(\mathrm{n}+1)(\mathrm{n}+2)}\right\}$ $\mathrm{T}_{1}=\frac{1}{2}\left\{\frac{1}{1.2}-\frac{1}{2.3}\right\}$ $\mathrm{T}_{2}=\frac{1}{2}\left\{\frac{1}{2.3}-\frac{1}{3.4}\right\}$ ⋮ $\mathrm{T}_{10}=\frac{1}{2}\left\{\frac{1}{10.11}-\frac{1}{11.12}\right\}$ $\mathrm{S}_{\mathrm{n}}=\frac{1}{2}\left[\frac{1}{2}-\frac{1}{132}\right]$ $\mathrm{S}_{\mathrm{n}}=\frac{1}{2}\left[\frac{66-1}{132}\right]=\frac{65}{264}$ So, $\lambda=\frac{65}{264} \times 528$ $\lambda=130$