_ n=1 ¹⁰ ( 528 n(n+1)(n+2) ) is equal to:

Step-by-step solution

Let T_n = 528 n(n+1)(n+2) T_n = 528 2 ( (n+2) - n n(n+1)(n+2) ) T_n = 264 ( 1 n(n+1) - 1 (n+1)(n+2) ) The sum of the first 10 terms is given by: S₁₀ = _ n=1 ¹⁰ T_n = 264 _ n=1 ¹⁰ ( 1 n(n+1) - 1 (n+1)(n+2) ) This is a telescoping series, so most terms cancel out: S₁₀ = 264 ( ( 1 1 2 - 1 2 3 ) + ( 1 2 3 - 1 3 4 ) + + ( 1 10 11 - 1 11 12 ) ) S₁₀ = 264 ( 1 2 - 1 11 12 ) S₁₀ = 264 ( 1 2 - 1 132 ) S₁₀ = 264 ( 66 - 1 132 ) S₁₀ = 264 65 132 S₁₀ = 2 65 = 130 Answer: 130