Find the limit of s(n) as n→ co. Part 1 of 3 s(n) = 81 n(n+1)(2n+1) 3 7 Apply the limit as n→ ∞ Part 2 of 3 lim n→ on both sides of s(n) = n- 81 n(n+1)(2n+1) 7 ∞ lim s(n) = n- ∞ ∞ [(31) n(n+1)(2n+1) 7 ∞ Now simplify lim_s(n) = lim n→ ∞ n→∞ ( 81 ) n (n + 81 n(n+1)(2n + 1) n° Submit lim s(n) lim 818 = noo (81) = (81) = [ ( 81 ) + 1) · 81 n (n + 1)(2n + 1) n° lim n∞ lim nco = ( 81 ) [ n(n+1)(2n+1) × +3 Skip (you cannot come back) to find the limit of s(n). +n Use the summation formulas to rewrite the expression without the summation notation. Use the result to find the sums for n = 10, 100, 1000, and 10,000. i = 1 2i + 6 n² Part 1 of 7 n Using the summation formula the expression 2i +6 n² can be written as the expression without i = 1 the summation notation. Therefore, Submit i = 1 2i + 6 n² 1 2 n i = 1 (2i + 6) 1 n 2 n Σ2+ ΣΕ i = 1 1 2 = 1 + = St יח 7nX n Skip (you cannot come back) 2 x ) 2 n(n + 1) + 6n ☑)