APPM 1350 Review Solutions 3

# 1 4n 2n 42 9 a write the sum in expanded form 10 x k6

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Unformatted text preview: e the sum 99 X ✓1 i=3 i 1 i+1 ◆ n X 1 ✓ j ◆2 (c) Evaluate the limits: lim , n!1 nn j =1 1 (d) lim n!1 n "✓ ◆ # ✓ ◆9 ⇣ n ⌘9 19 2 + + ··· + n n n Solution: 10 X (a) xk = x6 + x7 + x8 + x9 + x10 k=6 (b) 99 X ✓1 i=3 i 1 i+1 ◆ = = = ✓ ✓ 1 3 1 3 1 3 1 4 ◆ + ✓ 1 4 ◆✓ 1 1 ⇤⇤ + ⇤⇤ 4 4 ⇤ ⇤ 1 97 = 100 300 1 5 1 ⇤⇤ 5 ⇤ ◆ ◆ + + ✓ ✓ 1 5 1 6 1 ⇤⇤ 5 ⇤ 1 ⇤⇤ 6 ⇤ ◆ ◆ + ··· + + ··· + ✓ ✓ 1 99 1 100 1 99 1 100 ◆ ◆ (c) n n X 1 ✓ j ◆2 1X2 1 n(n + 1)(2n + 1) 2n3 + (lower order terms) 2 lim = lim 3 j = lim 3 · = lim = 3 |{z} 6 n!1 n!1 n n!1 n n!1 nn 6 6n j =1 j =1 DOP (d) If we let [a, b] = [0, 1], then x = 1/n and xi = i/n, then "✓ ◆ # ✓ ◆9 Z1 n n ⇣ n ⌘9 X ✓ i ◆9 1 X 1 19 2 9 lim + + ··· + = lim = lim ( xi ) x = x9 dx n!1 n n!1 n n n n n n!1 0 i=1 i=1 that is, the limit, under the right circumstances, can be interpreted as the deﬁnite integral and so we can evaluate the limit by evaluating the inte...
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