Let S_k = 1 + 2 + 3 + + k and Q_n = S₂ S₂ - 1 S₃ S₃ - 1 S₄ S₄ - 1 S_n S_n - 1 , where k, n N . _ n Q_n =

Step-by-step solution

We know S_n = n(n+1) 2 . S_n S_n - 1 = n(n+1) 2 n(n+1) 2 - 1 = n^2 + n n^2 + n - 2 = n(n+1) (n+2)(n-1) = ( n n-1 ) ( n+1 n+2 ) . So Q_n can be written as the product of two terms: Q_n = ( 2 1 3 2 n n-1 ) ( 3 4 4 5 n+1 n+2 ) . In the first bracket, everything cancels out except n 1 = n . In the second bracket, everything cancels out except 3 n+2 . Therefore, Q_n = n 3 n+2 = 3n n+2 . Taking the limit as n : _ n Q_n = _ n 3 1 + 2 n = 3 .