Summations, Inequalities, And Exponential Induction Quiz

1) To prove 1+2+…+n = n(n+1)/2, what is the statement P(k)?

1+2+…+k = k(k+1)/2

1+2+…+n = n(n+1)/2

1+2+…+(k+1) = (k+1)(k+2)/2

1 = 1(2)/2

In proof by induction, P(k) is the statement we assume to be true for an arbitrary integer k. For the summation formula, P(k) is the equation with n replaced by k, so it is 1+2+...+k = k(k+1)/2. This assumption is used in the inductive step to prove P(k+1).

Explanation

In proof by induction, P(k) is the statement we assume to be true for an arbitrary integer k. For the summation formula, P(k) is the equation with n replaced by k, so it is 1+2+...+k = k(k+1)/2. This assumption is used in the inductive step to prove P(k+1).

2) In proving 1+2+…+n = n(n+1)/2, what is the goal P(k+1)?

1+2+...+k = k(k+1)/2

1+2+...+k + (k+1) = (k+1)(k+2)/2

K+1 = k+1

The formula works for n=1

The goal of the inductive step is to prove that the statement holds for k+1, which means showing that the sum of the first k+1 integers equals (k+1)(k+2)/2. This is derived by adding the (k+1)th term to both sides of P(k) and simplifying to match the formula for n=k+1.

Explanation

The goal of the inductive step is to prove that the statement holds for k+1, which means showing that the sum of the first k+1 integers equals (k+1)(k+2)/2. This is derived by adding the (k+1)th term to both sides of P(k) and simplifying to match the formula for n=k+1.

3) To prove 3^n - 1 is divisible by 2, what is the base case for n ≥ 1?

3^n - 1 is divisible by 2

3^1 - 1 is divisible by 2

3^2 - 1 is divisible by 2

3^k - 1 is divisible by 2

The base case is the smallest value of n for which the statement is claimed to be true. Here, n≥1, so we test n=1: 3^1 - 1 = 2, which is divisible by 2. This establishes the initial truth needed for induction.

Explanation

The base case is the smallest value of n for which the statement is claimed to be true. Here, n≥1, so we test n=1: 3^1 - 1 = 2, which is divisible by 2. This establishes the initial truth needed for induction.

4) In an inductive proof, we see the line “Assume 3^k - 1 is divisible by 2. This means 3^k - 1 = 2m for some integer m.” What is the next step?

Prove the base case

Examine the expression for 3^{k+1} - 1

Assume 3^{k+1} - 1 is divisible by 2

Conclude the proof is complete

After stating the inductive hypothesis that 3^k - 1 is divisible by 2, the next step is to consider the expression for the next case: 3^{k+1} - 1. We rewrite this expression using algebraic manipulations, such as 3^{k+1} - 1 = 3 * 3^k - 1, and then use the inductive hypothesis to show it is also divisible by 2.

Explanation

After stating the inductive hypothesis that 3^k - 1 is divisible by 2, the next step is to consider the expression for the next case: 3^{k+1} - 1. We rewrite this expression using algebraic manipulations, such as 3^{k+1} - 1 = 3 * 3^k - 1, and then use the inductive hypothesis to show it is also divisible by 2.

5) To prove a summation formula, the inductive step usually involves:

Adding the (k+1)th term to both sides

Ignoring the base case

Using proof by contradiction

Subtracting the first term

For summation formulas, the inductive step often involves taking the assumed true statement P(k), which is the sum up to k, and then adding the (k+1)th term to both sides. This allows us to manipulate the left side to become the sum up to k+1 and the right side to simplify to the formula for k+1, thus proving P(k+1).

Explanation

For summation formulas, the inductive step often involves taking the assumed true statement P(k), which is the sum up to k, and then adding the (k+1)th term to both sides. This allows us to manipulate the left side to become the sum up to k+1 and the right side to simplify to the formula for k+1, thus proving P(k+1).

6) Proving P(1), P(2), and P(3) is a valid proof by induction.

True

False

Proof by induction requires proving the base case and the inductive step for all n. Proving only a finite number of cases (like P(1), P(2), P(3)) does not guarantee the statement for all natural numbers. The inductive step must show that if P(k) is true, then P(k+1) is true for all k, ensuring the statement holds indefinitely.

Explanation

Proof by induction requires proving the base case and the inductive step for all n. Proving only a finite number of cases (like P(1), P(2), P(3)) does not guarantee the statement for all natural numbers. The inductive step must show that if P(k) is true, then P(k+1) is true for all k, ensuring the statement holds indefinitely.

7) To prove 7^n - 1 is divisible by 6, we assume 7^k - 1 = 6m. How can we rewrite 7^{k+1}?

7 * 7^k

7 + 7^k

(7^k)^7

7^k

Using exponent rules, 7^{k+1} can be rewritten as 7 * 7^k. This is a common step in inductive proofs involving exponents, as it allows us to express the k+1 case in terms of the k case, facilitating the use of the inductive hypothesis.

Explanation

Using exponent rules, 7^{k+1} can be rewritten as 7 * 7^k. This is a common step in inductive proofs involving exponents, as it allows us to express the k+1 case in terms of the k case, facilitating the use of the inductive hypothesis.

8) In an inductive proof for an inequality like n < 2^n, the inductive step assumes:

K+1 < 2^{k+1}

1 < 2^1

K < 2^k

N < 2^n is always false

In the inductive step for an inequality, we assume the statement is true for an arbitrary k, i.e., k

Explanation

In the inductive step for an inequality, we assume the statement is true for an arbitrary k, i.e., k

9) If you prove a property for a sequence defined by a recurrence relation, you are using:

A direct proof

Proof by induction

Proof by counterexample

A geometric proof

Recurrence relations define sequences based on previous terms, making induction a natural proof technique. By proving the base case and showing that if the property holds for previous terms it holds for the next, we can establish the property for all terms in the sequence.

Explanation

Recurrence relations define sequences based on previous terms, making induction a natural proof technique. By proving the base case and showing that if the property holds for previous terms it holds for the next, we can establish the property for all terms in the sequence.

10) To prove 1^2 + ... + n^2 = n(n+1)(2n+1)/6, the term for k+1 is:

K^2

(k+1)^2

N^2

(k-1)^2

In the summation of squares, the (k+1)th term is (k+1)^2. When proving P(k+1), we add this term to the sum up to k, which is assumed to equal k(k+1)(2k+1)/6, and then show that the new sum equals (k+1)(k+2)(2k+3)/6.

Explanation

In the summation of squares, the (k+1)th term is (k+1)^2. When proving P(k+1), we add this term to the sum up to k, which is assumed to equal k(k+1)(2k+1)/6, and then show that the new sum equals (k+1)(k+2)(2k+3)/6.

11) Strong induction differs from standard (weak) induction in its:

Base case

Conclusion

Inductive hypothesis

Use of variables

In weak induction, the inductive hypothesis assumes P(k) is true for some k. In strong induction, the inductive hypothesis assumes that P(1), P(2), ..., P(k) are all true. This stronger assumption is useful when proving P(k+1) requires the truth of multiple previous cases.

Explanation

In weak induction, the inductive hypothesis assumes P(k) is true for some k. In strong induction, the inductive hypothesis assumes that P(1), P(2), ..., P(k) are all true. This stronger assumption is useful when proving P(k+1) requires the truth of multiple previous cases.

12) The inductive hypothesis for strong induction is:

Assume P(k) is true

Assume P(1), P(2), ..., P(k) are all true

Assume P(k+1) is true

Assume P(k-1) is true

Strong induction allows us to assume that the statement holds for all integers from the base case up to k, not just for k. This is particularly helpful for recurrence relations where the k+1 term depends on several previous terms.

Explanation

Strong induction allows us to assume that the statement holds for all integers from the base case up to k, not just for k. This is particularly helpful for recurrence relations where the k+1 term depends on several previous terms.

13) Any proof done with weak induction can also be done with strong induction.

True

False

Since strong induction has a stronger inductive hypothesis (it assumes more), any proof that can be done with weak induction can be rewritten using strong induction by simply not using the extra assumptions. However, the converse is not always true, as some proofs require the stronger hypothesis.

Explanation

Since strong induction has a stronger inductive hypothesis (it assumes more), any proof that can be done with weak induction can be rewritten using strong induction by simply not using the extra assumptions. However, the converse is not always true, as some proofs require the stronger hypothesis.

14) For a proof using strong induction, you may need to verify:

No base cases at all

Only one base case

Multiple base cases

The last case, P(n)

In strong induction, especially when the inductive step relies on several previous cases, it is often necessary to verify multiple base cases. For example, if P(k+1) depends on P(k-1) and P(k), then we need base cases for n=1 and n=2 to ensure the inductive step can be applied for all n.

Explanation

In strong induction, especially when the inductive step relies on several previous cases, it is often necessary to verify multiple base cases. For example, if P(k+1) depends on P(k-1) and P(k), then we need base cases for n=1 and n=2 to ensure the inductive step can be applied for all n.

15) Induction is a valid proof technique because of which axiom?

The Well-Ordering Principle

The Axiom of Choice

The Parallel Postulate

The Law of Syllogism

The Well-Ordering Principle states that every non-empty set of natural numbers has a least element. This principle is logically equivalent to the principle of mathematical induction. Induction works because if a statement is true for the base case and the inductive step, then the set of numbers where it is true must be all natural numbers, as otherwise there would be a smallest counterexample, contradicting the inductive step.

Explanation

The Well-Ordering Principle states that every non-empty set of natural numbers has a least element. This principle is logically equivalent to the principle of mathematical induction. Induction works because if a statement is true for the base case and the inductive step, then the set of numbers where it is true must be all natural numbers, as otherwise there would be a smallest counterexample, contradicting the inductive step.