MATH Section 5.1

Let P(n) be the statement that 13+ 23+ 33+ ...+ n313⁢+ 23⁢+ 33⁢+ ...⁢+ n3 = (n(n+ 1)2)2(n⁢(n⁢+ 1)2)2 for the positive integer n. The conditional statement P(k) → P(k + 1) is true for all positive integers k is called the inductive hypothesis. True or False Explanation: We assume that P(k) is true for an arbitrary positive integer k and show that under this assumption, P(k + 1) must also be true. The assumption that P(k) is true is called the inductive hypothesis.

False

et P(n) be the statement that 11⋅211⋅2 + 12⋅312⋅3 + ... + 1n⋅ (n+ 1)1n⋅ (n⁢+ 1) = n(n+1)n(n+1) . We will have completed the basis step of the proof if we show that (Check all that apply.) (You must provide an answer before moving to the next part.) Check All That Apply P(1) is true. P(0) is true. P(1) → P(2) is true. 1/2=1/2 11⋅2+12⋅3+...+1n(n+1)=nn+111⋅2+12⋅3+...+1n(n+1)=nn+1 is true for n = 1. 11⋅2+12⋅3+...+1n(n+1)=nn+111⋅2+12⋅3+...+1n(n+1)=nn+1 is true for some integer n. Explanation: We will have completed the basis step of the proof if we show P(1) is true because this is the basis step and 11⋅2+12⋅3+...+1n(n+1)=nn+111⋅2+12⋅3+...+1n(n+1)=nn+1 is true for n = 1 because this is what the notation P(1) means.

P(1) is true. 11⋅2+12⋅3+...+1n(n+1)=nn+111⋅2+12⋅3+...+1n(n+1)=nn+1 is true for n = 1.

Let P(n) be the statement n2 ≤ n! where n is a nonnegative integer. For which nonnegative integers n is P(n)? (You must provide an answer before moving to the next part.) Multiple Choice n = 0, n = 1, n ≥ 4 n = 1, n = 2, n = 4 n = 2, n ≥ 4 n = 3, n ≥ 4 n = 0, n = 2, n > 4 Explanation: P(n) appears to be true for n = 0, n = 1, n ≥ 4. (Clearly, the inequality does not hold for n = 2 or n = 3.)

n = 0, n = 1, n ≥ 4

Let P(n) be the statement that 11⋅211⋅2 + 12⋅312⋅3 + ... + 1n⋅ (n+ 1)1n⋅ (n⁢+ 1) = n(n+1)n(n+1) . Identify the statement P(k + 1). Multiple Choice 11⋅2+12⋅3+...+1k⋅(k+1)+1(k+1)(k−1)=k+1k+211·2+12·3+...+1k·(k+1)+1(k+1)(k-1)=k+1k+2 is true for a specific k > 0. 11⋅2+12⋅3+...+1k⋅(k+1)+1(k−1)(k+2)=k+1k+211·2+12·3+...+1k·(k+1)+1(k-1)(k+2)=k+1k+2 is true for a specific k < 0. 11⋅2+12⋅3+...+1k⋅(k+1)+1(k+1)(k−2)=k+1k+211·2+12·3+...+1k·(k+1)+1(k+1)(k-2)=k+1k+2 is true for a specific k < 0. 11⋅2+12⋅3+...+1k⋅(k+1)+1(k+1)(k+2)=k+1k+211·2+12·3+...+1k·(k+1)+1(k+1)(k+2)=k+1k+2 is true for a specific k > 0. Explanation: Let P(k) be the statement that 11⋅2+12⋅3+...+1k(k+1)=kk+111·2+12·3+...+1k(k+1)=kk+1 . P(k+ 1) is 11⋅2+12⋅3+...+1k⋅(k+1)+1(k+1)(k+2)=k+1k+211·2+12·3+...+1k·(k+1)+1(k+1)(k+2)=k+1k+2is true for a specifick> 0 as desired.

11⋅2+12⋅3+...+1k⋅(k+1)+1(k+1)(k+2)=k+1k+211·2+12·3+...+1k·(k+1)+1(k+1)(k+2)=k+1k+2 is true for a specific k > 0.

Let P(n) be the statement n2 ≤ n! where n is a nonnegative integer. Click and drag the steps given (in the right) to their corresponding step names (in the left) to prove that n2 ≤ n!n2⁢ ≤ n! for all n ≥ 4 using mathematical induction. Explanation: We will prove by mathematical induction that the inequality holds for all n ≥ 4. The basis step is clear since 16 ≤ 24. Now suppose that k2 ≤ k! for a given k ≥ 4. We must show that (k + 1)2 ≤ (k + 1)!. Expanding the left-hand side, applying the inductive hypothesis, and then invoking some valid bounds shows the following: k2 + 2k + 1 ≤ k! + 2k + 1≤ k! + 2k + k = k! + 3k≤ k! + k · k ≤ k! + k · k!= (k + 1)k! = (k + 1)!

Correct blocks and order: Basis Step: We will prove by mathematical induction that the inequality holds for all n ≥ 4. Since 16 ≤ 24. Inductive hypothesis: Now suppose that k2 < k! for a given k ≥ 4. We must show that (k + 1)2 ≤ (k + 1)! Inductive step, part 1: Expanding the left hand side, and applying the inductive hypothesis, we get Inductive step, part 2: ≤ k! + 2k + k = k! + 3k Inductive step, part 3: ≤ k! + k ⋅ k ≤ k! + k ⋅ k! Inductive step, part 4: = (k + 1)k! = (k + 1)!

Let P(n) be the statement that 13+ 23+ 33+ ...+ n313⁢+ 23⁢+ 33⁢+ ...⁢+ n3 = (n(n+ 1)2)2(n⁢(n⁢+ 1)2)2 for the positive integer n. What do you need to prove in the inductive step? (You must provide an answer before moving to the next part.) Multiple Choice If 13+23+⋯+k3=(k(k+1)2)213+23+⋯+k3=k(k+1)22 , then 13 + 23 + ...+ k313⁢ + 23⁢ + ...⁢+ k3 + (k+ 1)3(k⁢+ 1)3 = (k (k+ 1)2)2(k⁢ (k⁢+ 1)2)2 . If 13+23+⋯+k3=(k(k+1)2)213+23+⋯+k3=k(k+1)22 , then 13 + 23 + ...+ k313⁢ + 23⁢ + ...⁢+ k3 = ((k+ 1)(k+ 2)2)2((k⁢+ 1)(k⁢+ 2)2)2 . If 13+23+⋯+k3=(k(k+1)2)213+23+⋯+k3=k(k+1)22 , then 13 + 23 + ...+ k313⁢ + 23⁢ + ...⁢+ k3 + (k+ 1)3(k⁢+ 1)3 = ((k+ 1)(k+ 2)2)3((k⁢+ 1)(k⁢+ 2)2)3 . If 13+23+⋯ +k3=(k(k+1)2)213+23+⋯ +k3=k(k+1)22 , then 13 + 23 + ...+ k313⁢ + 23⁢ + ...⁢+ k3 + (k+ 1)3(k⁢+ 1)3 = ((k+ 1)(k+ 2)2)2((k⁢+ 1)(k⁢+ 2)2)2 . Explanation: For the inductive hypothesis, we assume that P(k) holds for an arbitrary positive integer k. That is, we assume that 13+23+⋯+k3=(k(k+1)2)213+23+⋯+k3=k(k+1)22 Under this assumption, it must be shown that P(k + 1) is true, namely, that 13+23+⋯+k3+(k+1)3=((k+1)(k+2)2)2

If 13+23+⋯ +k3=(k(k+1)2)213+23+⋯ +k3=k(k+1)22 , then 13 + 23 + ...+ k313⁢ + 23⁢ + ...⁢+ k3 + (k+ 1)3(k⁢+ 1)3 = ((k+ 1)(k+ 2)2)2((k⁢+ 1)(k⁢+ 2)2)2 .

et P(n) be the statement that 11⋅211⋅2 + 12⋅312⋅3 + ... + 1n⋅ (n+ 1)1n⋅ (n⁢+ 1) = n(n+1)n(n+1) . Identify the inductive hypothesis. (You must provide an answer before moving to the next part.) Multiple Choice In the inductive hypothesis, we assume P(k) for some integer k, k > 0, 11⋅211⋅2 + 12⋅312⋅3 + ... + 1k⋅ (k+ 1)1k⋅ (k⁢+ 1) = k− 1k+ 1 k⁢− 1k⁢+ 1 . In the inductive hypothesis, we assume P(k) for some integer k, k > 0, 11⋅211⋅2 + 12⋅312⋅3 + ... + 1k⋅ (k+ 1)1k⋅ (k⁢+ 1) = k+1k+2k+1k⁢+2 . In the inductive hypothesis, we assume P(k) for some integer k, k > 0, 11⋅211⋅2 + 12⋅312⋅3 + ... + 1k⋅ (k+ 1)1k⋅ (k⁢+ 1) = kk+ 1kk⁢+ 1 . In the inductive hypothesis, we assume P(k) for some integer k, k < 0, 11⋅211⋅2 + 12⋅312⋅3 + ... + 1k⋅ (k+ 1)1k⋅ (k⁢+ 1) = k−1k+ 1k−1k⁢+ 1 . Explanation: We assume that P(k) is true for an arbitrary positive integer k. The assumption that P(k) is true is called the inductive hypothesis. In the inductive hypothesis, we assume P(k) for some integer k, k > 0, 11⋅211⋅2 + 12⋅312⋅3 + ... + 1k⋅ (k+ 1)1k⋅ (k⁢+ 1) = kk+ 1kk⁢+ 1 .

In the inductive hypothesis, we assume P(k) for some integer k, k > 0, 11⋅211⋅2 + 12⋅312⋅3 + ... + 1k⋅ (k+ 1)1k⋅ (k⁢+ 1) = kk+ 1kk⁢+ 1 .

Let P(n) be the statement that 13+ 23+ 33+ ...+ n313⁢+ 23⁢+ 33⁢+ ...⁢+ n3 = (n(n+ 1)2)2(n⁢(n⁢+ 1)2)2 for the positive integer n. We will have completed the basis step of the proof if we show that (Check all that apply.) Check All That Apply P(1) is true. P(1) is true. Correct P(0) is true. P(0) is true. Correct P(1) → P(2) is true. P(1) → P(2) is true. Correct 13+23+...+n3=(n(n+1)2)213+23+...+n3=(n(n+1)2)2 is true for n = 1. 1 3 + 2 3 +... + n3 = ( n(n+1 )2 )2 is true for n = 1. Correct 13+23+...+n3=(n(n+1)2)213+23+...+n3=(n(n+1)2)2 is true for some integer n. Explanation: We will have completed the basis step of the proof if we show P(1) is true because this is the basis step and 13+23+...+n3=(n(n+1)2)213+23+...+n3=(n(n+1)2)2 is true for n = 1 because this is what the notation P(1) means. i.e., 13=(1(1+1)2)2=113=1(1+1)22=1 . (The left-hand side of this equation is 1 because 1 is the sum of the first positive integer. The right-hand side is found by substituting 1 for n in n(n + 1)/2.)

P(1) is true. 13+23+...+n3=(n(n+1)2)213+23+...+n3=(n(n+1)2)2 is true for n = 1.

Consider the statement that 3 divides n3 + 2n whenever n is a positive integer. Outline the proof by clicking and dragging to complete each statement. Click and drag statements to fill in the details of showing that ∀∀ k(P(k) →→ P(k + 1)) is true, thereby completing the induction step. Explanation: Let P(n) be the proposition that 3 divides n3 + 2n. P(1) states that 3 divides 13 + 2(1) = 3, and 3 divides 3. Assume that 3 divided k3 + 2k for any arbitrary integer k > 0. So, we have to show that ∀k > 0, (P(k) → P(k + 1)) is true, that is, ∀k > 0, (3 divides k3 + 2k → 3 divides (k + 1)3 + 2(k + 1)). By the inductive hypothesis, 3 divides (k3 + 2k). 3 divides 3(k2 + k + 1) because it is 3 times an integer. By the theorem "if a divides b and a divides c, then a divides b + c," 3 divides the sum (k3 + 2k) +3(k2 + k + 1). This completes the inductive step. We have completed the basis step and the inductive step. By mathematical induction, we know that P(n) is true for all integers n ≥ 1.

Part 1: Let P(n) be the proposition that: 3 divides n^3 + 2n. Basis step: P(1) states that: 3 divides 13 + 2 ⋅ 1, which is true since 13 + 2 ⋅ 1 = 3, and 3 divides 3. Inductive step: Assume that 3 divided k3 + 2k for an arbitrary integer k > 0. Show that ∀k > 0, (P(k) → P(k + 1)) is true, that is ∀k > 0, (3 divides k3 + 2k → 3 divides (k + 1)3 + 2(k + 1)). We have completed the basis step and the inductive step. By mathematical induction, we know that P(n) is true for all integers n ≥ 1. Part 2: steps (order going down): (k + 1)3 + 2(k + 1) = (k3 + 3k2 + 3k + 1) + (2k + 2) = (k3 + 2k) + 3(k2 + k + 1) By the inductive hypothesis, 3 divides (k3 + 2k). 3 divides 3(k2 + k + 1) because it is 3 times an integer. By part (i) of Theorem 1 in Section 4.1, 3 divides the sum (k3 + 2k) + 3(k2 + k + 1). This completes the inductive step.

Let P(n) be the statement that 13+ 23+ 33+ ...+ n313⁢+ 23⁢+ 33⁢+ ...⁢+ n3 = (n(n+ 1)2)2(n⁢(n⁢+ 1)2)2 for the positive integer n. Complete the inductive step, identifying where you use the inductive hypothesis. (You must provide an answer before moving to the next part.) Multiple Choice Replacing the quantity in brackets on the left-hand side of part (c) by what it equals by virtue of the inductive hypothesis, we have (k(k+1)2)3+(k+1)2=(k+1)2(k2+4k+44)2=((k+1)(k+2)4)2(k(k+1)2)3+(k+1)2=(k+1)2(k2+4k+44)2=((k+1)(k+2)4)2 as desired. Replacing the quantity in brackets on the left-hand side of part (c) by what it equals by virtue of the inductive hypothesis, we have (k(k+1)2)2+(k+1)2=(k+1)2(k2+4k+14)=((k+1)(k+2)4)2(k(k+1)2)2+(k+1)2=(k+1)2(k2+4k+14)=((k+1)(k+2)4)2 as desired. Replacing the quantity in brackets on the left-hand side of part (c) by what it equals by virtue of the inductive hypothesis, we have (k(k+1)2)2+(k+1)3=(k+1)2(k2+4k+44)=((k+1)(k+2)2)2(k(k+1)2)2+(k+1)3=(k+1)2(k2+4k+44)=((k+1)(k+2)2)2 as desired. Replacing the quantity in brackets on the left-hand side of part (c) by what it equals by virtue of the inductive hypothesis, we have (k(k+1)2)2+(k+1)3=(k+1)3(k2+4k+14)=((k+1)(k+2)2)2(k(k+1)2)2+(k+1)3=(k+1)3(k2+4k+14)=((k+1)(k+2)2)2 as desired. Explanation: For the inductive hypothesis, we assume that P(k) holds for an arbitrary positive integer k. That is, we assume that 13+23+⋯+k3=(k+12)213+23+⋯+k3=k+122 . Under this assumption, it must be shown that P(k + 1) is true, namely, that 13+23+⋯+k3+(k+1)3=((k+1)(k+2)2)213+23+⋯+k3+(k+1)3=(k+1)(k+2)22 . Replacing the quantity in brackets on the left-hand side of the above statement by what it equals by virtue of the inductive hypothesis, we have (k(k+1)2)2+(k+1)3=(k+1)2(k2+4k+44)=((k+1)(k+2)2)2(k(k+1)2)2+(k+1)3=(k+1)2(k2+4k+44)=((k+1)(k+2)2)2 as desired.

Replacing the quantity in brackets on the left-hand side of part (c) by what it equals by virtue of the inductive hypothesis, we have (k(k+1)2)2+(k+1)3=(k+1)2(k2+4k+44)=((k+1)(k+2)2)2(k(k+1)2)2+(k+1)3=(k+1)2(k2+4k+44)=((k+1)(k+2)2)2 as desired.

Let P(n) be the statement that 13+ 23+ 33+ ...+ n313⁢+ 23⁢+ 33⁢+ ...⁢+ n3 = (n(n+ 1)2)2(n⁢(n⁢+ 1)2)2 for the positive integer n. Explain why these steps show that this formula is true whenever n is a positive integer. Multiple Choice We have completed both the basis step and the inductive step, so by the principle of mathematical induction, the statement is true for every real number n. We have completed both the basis step and the inductive step, so by the principle of mathematical induction, the statement is true for every integer n. We have completed the inductive step, so by the principle of mathematical induction, the statement is true for every positive integer n. We have completed both the basis step and the inductive step, so by the principle of mathematical induction, the statement is true for every positive integer n. We have completed the inductive step, so by the principle of mathematical induction, the statement is true for every integer n. Explanation: Both the basis step and the inductive step are true for the given statement. By the principle of mathematical induction, the given statement is true for every positive integer n.

We have completed both the basis step and the inductive step, so by the principle of mathematical induction, the statement is true for every positive integer n.

Arrange the given steps in the correct order to prove that 11⋅211⋅2 + 12⋅312⋅3 + ... + 1k⋅ (k+ 1)1k⋅ (k⁢+ 1) = kk+1kk⁢+1 is true for all n > 0 using the concept of mathematical induction. Explanation: For n = 1, the left-hand side of the theorem is 11⋅211⋅2 = 1212 and the right-hand side = nn + 1nn⁢ + 1 = 1212 . Assume that for some k > 0, 11⋅211⋅2 + 12⋅312⋅3 + ... + 1k⋅ (k+ 1)1k⋅ (k⁢+ 1) = kk+1kk⁢+1 . By adding 1(k+1)(k+2)1(k+1)(k+2) on both the sides, we get 11⋅211⋅2 + 12⋅312⋅3 + ... + 1k⋅ (k+ 1)1k⋅ (k⁢+ 1) + 1(k+ 1)(k + 2)1(k⁢+ 1)(k⁢ + 2) = kk+ 1kk⁢+ 1 + 1(k+ 1)(k + 2)1(k⁢+ 1)(k⁢ + 2) . Therefore, we have 11⋅2+12⋅3+...+1k⋅(k+1)+1(k+1)(k+2)=k(k+ 2) + 1(k+ 1)(k+ 2)11⋅2+12⋅3+...+1k⋅(k+1)+1(k+1)(k+2)=k(k⁢+ 2) + 1(k⁢+ 1)(k⁢+ 2) . We have completed both the basis step and the inductive step; so, by the principle of mathematical induction, the statement is true for every positive integer n.

steps: For n = 1, the left-hand side of the theorem is 11⋅211⋅2 = 1212 and the right-hand side = nn + 1nn⁢ + 1 = 1212 . Assume that for some k > 0, 11⋅211⋅2 + 12⋅312⋅3 + ... + 1k⋅ (k+ 1)1k⋅ (k⁢+ 1) = kk+1kk⁢+1 . By adding 1(k+1)(k+2)1(k+1)(k+2) on both the sides, we get 11⋅211⋅2 + 12⋅312⋅3 + ... + 1k⋅ (k+ 1)1k⋅ (k⁢+ 1) + 1(k+ 1)(k + 2)1(k⁢+ 1)(k⁢ + 2) = kk+ 1kk⁢+ 1 + 1(k+ 1)(k + 2)1(k⁢+ 1)(k⁢ + 2) . Therefore, we have 11⋅2+12⋅3+...+1k⋅(k+1)+1(k+1)(k+2)=k(k+ 2) + 1(k+ 1)(k+ 2)11⋅2+12⋅3+...+1k⋅(k+1)+1(k+1)(k+2)=k(k⁢+ 2) + 1(k⁢+ 1)(k⁢+ 2) . This is true because each term on left hand-side = kk+1kk+1 , by the inductive hypothesis. Therefore, k(k+2)+1(k+1)(k+2)= k2+2k+1(k+1)(k+2)=k+1k+2k(k+2)+1(k+1)(k+2)= k2+2k+1(k+1)(k+2)=k+1k+2 . We have completed both the basis step and the inductive step; so, by the principle of mathematical induction, the statement is true for every positive integer n.