Respuesta :
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
