About Prove each of the following statements using mathematical induction. (a) Prove that for any positive integer n, 4 evenly divides 32n-1. (b) Prove that for any positive integer n, 6 evenly divides 7n - 1. (c) Prove that for any positive integer n, 4 evenly divides 11n - 7n. (d) Prove that for any positive integer n, 7 evenly divides 9n - 2n.

Answers

Answer 1

Answer:

(a) [tex]3^{2{k+1}} - 1 = 4(9k + 2)[/tex]

(b) [tex]7^{k+1} - 1 = 6(7k + 1)[/tex]

(c) [tex]11(k + 1) - 7(k + 1) = 4(k + 1)[/tex]

(d) [tex]9(k+1) - 2(k+1) = 7(k+1)[/tex]

Step-by-step explanation:

Solving (a): For integer [tex]n[/tex], 4 divides [tex]3^{2n} - 1[/tex]

The proof is as follows:

For n = k, we have:

[tex]\frac{3^{2k}- 1}{4} = k[/tex]

Multiply through by 4

[tex]3^{2k} - 1 = 4k[/tex]

[tex]9^k - 1 = 4k[/tex]

Next, prove the statement is true for [tex]n = k+1[/tex]

[tex]3^{2{k+1}} - 1[/tex]

[tex]3^{2{k+1}} - 1 = 9^{{k+1}} - 1[/tex]

[tex]3^{2{k+1}} - 1 = 9^{k} * 9^1 - 1[/tex]

[tex]3^{2{k+1}} - 1 = 9^k * 9 - 1[/tex]

Express -1 as - 9 + 8

[tex]3^{2{k+1}} - 1 = 9^k * 9 - 9 + 8[/tex]

Factorize:

[tex]3^{2{k+1}} - 1 = 9(9^k - 1) + 8[/tex]

Recall that: [tex]9^k - 1 = 4k[/tex]

So, we have:

[tex]3^{2{k+1}} - 1 = 9*4k + 8[/tex]

Factorize

[tex]3^{2{k+1}} - 1 = 4(9k + 2)[/tex]

Since the above mathematical statement is true, then the given statement has been proved

Solving (b): For integer [tex]n[/tex], 6 divides [tex]7^n - 1[/tex]

The proof is as follows:

For n = k, we have:

[tex]\frac{7^k - 1}{6} = k[/tex]

Multiply through by 6

[tex]7^k - 1 = 6k[/tex]

Next, prove the statement is true for [tex]n = k+1[/tex]

[tex]7^{k+1} - 1[/tex]

[tex]7^{k+1} - 1 = 7^k * 7^1 - 1[/tex]

[tex]7^{k+1} - 1 = 7^k * 7 - 1[/tex]

Express -1 as - 7 + 6

[tex]7^{k+1} - 1 = 7^k * 7 - 7 + 6[/tex]

Factorize:

[tex]7^{k+1} - 1 = 7(7^k - 1) + 6[/tex]

Recall that: [tex]7^k - 1 = 6k[/tex]

So, we have:

[tex]7^{k+1} - 1 = 7(6k) + 6[/tex]

Factorize

[tex]7^{k+1} - 1 = 6(7k + 1)[/tex]

Since the above mathematical statement is true, then the given statement has been proved

Solving (c): For integer [tex]n[/tex], 4 divides [tex]11n - 7n[/tex]

The proof is as follows:

For n = k, we have:

[tex]\frac{11k - 7k}{4} = k[/tex]

Multiply through by 4

[tex]11k - 7k = 4k[/tex]

Next, prove the statement is true for [tex]n = k+1[/tex]

[tex]11(k + 1) - 7(k + 1)[/tex]

[tex]11(k + 1) - 7(k + 1) = 11k + 11 - 7k - 7[/tex]

Collect Like Terms

[tex]11(k + 1) - 7(k + 1) = 11k - 7k + 11 - 7[/tex]

[tex]11(k + 1) - 7(k + 1) = 11k - 7k + 4[/tex]

Recall that: [tex]11k - 7k = 4k[/tex]

So, we have:

[tex]11(k + 1) - 7(k + 1) = 4k + 4[/tex]

Factorize

[tex]11(k + 1) - 7(k + 1) = 4(k + 1)[/tex]

Since the above mathematical statement is true, then the given statement has been proved

Solving (d): For integer [tex]n[/tex], 7 divides [tex]9n - 2n[/tex]

The proof is as follows:

For n = k, we have:

[tex]\frac{9k - 2k}{7} = k[/tex]

Multiply through by 7

[tex]9k - 2k = 7k[/tex]

Next, prove the statement is true for [tex]n = k+1[/tex]

[tex]9(k+1) - 2(k+1)[/tex]

[tex]9(k+1) - 2(k+1) = 9k+9 - 2k-2[/tex]

Collect Like Terms

[tex]9(k+1) - 2(k+1) = 9k- 2k+9 -2[/tex]

[tex]9(k+1) - 2(k+1) = 9k- 2k+7[/tex]

Recall that: [tex]9k - 2k = 7k[/tex]

So, we have:

[tex]9(k+1) - 2(k+1) = 7k+7[/tex]

Factorize

[tex]9(k+1) - 2(k+1) = 7(k+1)[/tex]

Since the above mathematical statement is true, then the given statement has been proved


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Step-by-step explanation:

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Step-by-step explanation:

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What are equation models?

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here,
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Answer:

Step-by-step explanation:

Answer:

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Answer:

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Here is the answer. Solved

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This is the answer for the question

The proof is shown below:

What is complex number?

A complexb number is the of a real number and an imaginary number. A complex number is of the form a + ib and is usually represented by z. Here both a and b are real numbers. The value 'a' is called the real part which is denoted by Re(z), and 'b' is called the imaginary part Im(z).  Also, ib is called an imaginary number.

The alphabet i is referred to as the iota and is helpful to represent the imaginary part of the complex number. Further the iota(i) is very helpful to find the square root of negative numbers.

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Answers

Answer:

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Step-by-step explanation:

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Answers

Answer:

Kindly check explanation

Step-by-step explanation:

H0 : μ ≥ 402

H1 : μ < 402

Test statistic :

n = 23

Sample Mean, x = 400

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Test statistic = (x - μ) ÷ σ/sqrt(n)

Test statistic = (400 - 402) ÷ 30/sqrt(23)

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Critical value :

Using the Tcritical value calculator

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Tcritical = 2.074

Reject Null : if Test statistic ≤ Tcritical (left tail test)

Since ;

Test statistic ≤ Tcritical ; We reject the Null

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Answers

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Step-by-step explanation:

Step-by-step explanation:

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