𝒏(𝒏+𝟏)/2∈Θ(𝒏^3)
The statement n(n+1)/2 ∈ Θ(n^3) is not accurate because the growth behavior of n(n+1)/2 is not cubic. Instead, we can analyze the function's growth rate more accurately by proving it belongs to the Θ(n²) class.\n\nTo show that n(n+1)/2 ∈ Θ(n²), we need to demonstrate that there exist positive constants c1, c2, and n0 such that for all n >= n0, we have:\n\nc1 * n² <= n(n+1)/2 <= c2 * n²\n\nWe already proved the lower bound for Ω(n) earlier. Now we need to find the lower and upper bounds for Θ(n²). \n\nLower bound:\n\nn(n+1)/2 >= c1 * n²\n\nDividing both sides by n², we obtain:\n\n(n+1)/(2n) >= c1\n\nAs n approaches infinity, the term (n+1)/(2n) approaches 1/2. So, we can choose c1 = 1/2 and n0 = 1, satisfying the lower bound for all n >= 1.\n\nUpper bound:\n\nn(n+1)/2 <= c2 * n²\n\nDividing both sides by n², we get:\n\n(n+1)/(2n) <= c2\n\nAs n becomes arbitrarily large, the term (n+1)/(2n) converges to 1/2. Choosing c2 = 1/2 and n0 = 1 also satisfies the upper bound for all n >= 1.\n\nSince we have found positive constants c1, c2, and n0 that satisfy the lower and upper bounds for Θ(n²), we can conclude that n(n+1)/2 ∈ Θ(n²), not Θ(n^3).