5.1 MTH 288

NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. Suppose that n is a positive integer. Arrange the steps in the correct order to show that 2 divides n2 + n using mathematical induction.

1. 12 + 1 = 2 and 2 | 2, so the basis step is clear. 2. The inductive hypothesis is that 2 divides k2 + k. 3. (k + 1)2 + (k + 1) = (k2 + 2k + 1) + (k + 1) = (k2 + k ) + 2(k + 1) 4. 2 divides k2 + k by inductive hypothesis, and clearly 2 divides 2(k + 1). 5. As the sum of two multiples of 2 is again a multiple of 2, 2 divides (k + 1)2 + (k + 1).

NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. Let P(n) be the statement that 11⋅211⋅2 + 12⋅312⋅3 + ... + 1n⋅ (n+ 1)1�⋅ (�⁢+ 1) = n(n+1)�(�+1) . Identify the statement P(k + 1).

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

NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. Let P(n) be the statement that 13+ 23+ 33+ ...+ n313⁢+ 23⁢+ 33⁢+ ...⁢+ �3 = (n(n+ 1)2)2(�⁢(�⁢+ 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. (You must provide an answer before moving to the next part.)

False

NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. Let P(n) be the statement that 13+ 23+ 33+ ...+ n313⁢+ 23⁢+ 33⁢+ ...⁢+ �3 = (n(n+ 1)2)2(�⁢(�⁢+ 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.)

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

NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. Let P(n) be the statement that 11⋅211⋅2 + 12⋅312⋅3 + ... + 1n⋅ (n+ 1)1�⋅ (�⁢+ 1) = n(n+1)�(�+1) . Identify the inductive hypothesis. (You must provide an answer before moving to the next part.)

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

NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. Let P(n) be the statement that 11⋅211⋅2 + 12⋅312⋅3 + ... + 1n⋅ (n+ 1)1�⋅ (�⁢+ 1) = n(n+1)�(�+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.)

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

NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. Let P(n) be the statement that 13+ 23+ 33+ ...+ n313⁢+ 23⁢+ 33⁢+ ...⁢+ �3 = (n(n+ 1)2)2(�⁢(�⁢+ 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.) (You must provide an answer before moving to the next part.)

P(1) is true. 13+23+...+n3=(n(n+1)2)213+23+...+�3=(�(�+1)2)2 is true for n = 1.

NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. Let P(n) be the statement that 13+ 23+ 33+ ...+ n313⁢+ 23⁢+ 33⁢+ ...⁢+ �3 = (n(n+ 1)2)2(�⁢(�⁢+ 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.)

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(�(�+1)2)2+(�+1)3=(�+1)2(�2+4�+44)=((�+1)(�+2)2)2 as desired.

NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. Suppose that n is a positive integer. Identify the steps involved in proving that 2 divides n2 + n without using mathematical induction. (Check all that apply.)

We can express n2 + n as n(n + 1). As n and n + 1 are consecutive integers, one of them is even and the other is odd. Hence, the product n(n + 1) is even. Therefore, 2 divides n(n + 1), i.e., 2 divides n2 + n.

NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. Let P(n) be the statement that 13+ 23+ 33+ ...+ n313⁢+ 23⁢+ 33⁢+ ...⁢+ �3 = (n(n+ 1)2)2(�⁢(�⁢+ 1)2)2 for the positive integer n. Explain why these steps show that this formula is true whenever n is a positive integer.

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.

NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. 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!�2⁢ ≤ �! for all n ≥ 4 using mathematical induction.

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)!

NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. 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.)

n = 0, n = 1, n ≥ 4