Question

Sol. [Key Concept: Telescope $T_{n} = \frac{1}{2}(\frac{1}{n(n + 1)} - \frac{1}{(n + 1)(n + 2)})$, sum, then multiply by $528$.]

Let $\sum_{n = 1}^{10}\mspace{2mu}\frac{528}{n(n + 1)(n + 2)} = \lambda$

Let $T_{n} = \frac{1}{n(n + 1)(n + 2)}$

$${T_{n} = \frac{1}{2}\left\{ \frac{(n + 2) - n}{n(n + 1)(n + 2)} \right\} }{T_{n} = \frac{1}{2}\left\{ \frac{1}{n(n + 1)} - \frac{1}{(n + 1)(n + 2)} \right\} }{T_{1} = \frac{1}{2}\left\{ \frac{1}{1.2} - \frac{1}{2.3} \right\} }{T_{2} = \frac{1}{2}\left\{ \frac{1}{2.3} - \frac{1}{3.4} \right\} }$$⋮

$${T_{10} = \frac{1}{2}\left\{ \frac{1}{10.11} - \frac{1}{11.12} \right\} }{S_{n} = \frac{1}{2}\left\lbrack \frac{1}{2} - \frac{1}{132} \right\rbrack }{S_{n} = \frac{1}{2}\left\lbrack \frac{66 - 1}{132} \right\rbrack = \frac{65}{264} }$$So, $\lambda = \frac{65}{264} \times 528$

$$\lambda = 130 $$