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If the sum of the first n terms of a sequence is given by sum_(r=1)ⁿ T_r = n(n² - 1), find the value of sum_(r=2)^(infinity) 1/T_r.

1. 1/3
2. 1/2
3. 2/3
4. 1

Correct answer: 1/2

Solution

Sₙ = n(n²-1) = n(n-1)(n+1). Tₙ = Sₙ - S_(n-1) = n(n-1)(n+1) - (n-1)(n-2)n = n(n-1)[(n+1)-(n-2)] = n(n-1)*3 = 3n(n-1). Wait: S_(n-1) = (n-1)((n-1)² - 1) = (n-1)(n²-2n) = (n-1)n(n-2). Tₙ = n(n-1)(n+1) - (n-1)n(n-2) = n(n-1)[(n+1)-(n-2)] = n(n-1)*3 = 3n(n-1). For r >= 2: 1/T_r = 1/(3r(r-1)). Sum = (1/3) * sum_(r=2)^(inf) 1/(r(r-1)) = (1/3) * sum_(r=2)^(inf) (1/(r-1) - 1/r) = (1/3)*1 = 1/3.

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