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Click and drag the steps in the correct order to show that 3 divides n³ + 2n whenever n is a positive integer using mathematical induction. BASIS STEP: By the inductive hypothesis, 3 | (k³ + 2k), and certainly 3 | 3(k² + 1). As the sum of two multiples of 3 is again divisible by 3, 3 ((k+1)3 + 2(k+ 1)). INDUCTIVE STEP: 31 (03+2 0), i.e., 310, so the basis step is true. (k+1)+2(k+1)=(k + 3k² + 1) + (2k + 2) = (k³ + 2k) + 3(k² + 1) Suppose that 3|(k³ + 2k). (k+1)+2(k+1)=(k³ + 3k²+3k+ 1) + (2k+2)= (k³+2k)+3(k²+k+1) By the inductive hypothesis, 3 | (k³ + 2k), and certainly 3 | 3(k² + k + 1). 3|(1+2 1), i.e., 3|3, so the basis step is true.

Click and drag the steps in the correct order to show that 3 divides n³ + 2n whenever n is a positive integer using mathematical induction. BASIS STEP: By the inductive hypothesis, 3 | (k³ + 2k), and certainly 3 | 3(k² + 1). As the sum of two multiples of 3 is again divisible by 3, 3 ((k+1)3 + 2(k+ 1)). INDUCTIVE STEP: 31 (03+2 0), i.e., 310, so the basis step is true. (k+1)+2(k+1)=(k + 3k² + 1) + (2k + 2) = (k³ + 2k) + 3(k² + 1) Suppose that 3|(k³ + 2k). (k+1)+2(k+1)=(k³ + 3k²+3k+ 1) + (2k+2)= (k³+2k)+3(k²+k+1) By the inductive hypothesis, 3 | (k³ + 2k), and certainly 3 | 3(k² + k + 1). 3|(1+2 1), i.e., 3|3, so the basis step is true.

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Click and drag the steps in the correct order to show that 3 divides n³ + 2n whenever n is a positive integer using mathematical induction. BASIS STEP: By the inductive hypothesis, 3 | (k³ + 2k), and certainly 3 | 3(k² + 1). As the sum of two multiples of 3 is again divisible by 3, 3 | ((k+1)3 + 2(k + 1)). INDUCTIVE STEP: 31 (03+2 0), i.e., 3 | 0, so the basis step is true. (k+1)3 + 2(k+1) = (k³ + 3k² + 1) + (2k + 2) = (k³ + 2k) + 3(k² + 1) Suppose that 3 | (k³ + 2k). (k+1)3 + 2(k+1) = (k³ + 3k² + 3k+ 1) + (2k + 2) = (k³ +2k)+3(k²+k+1) By the inductive hypothesis, 3 | (k³ + 2k), and certainly 3 | 3(k² + k + 1). 3|(13+2 1), i.e., 3 | 3, so the basis step is true. Reset
Transcribed Image Text:Click and drag the steps in the correct order to show that 3 divides n³ + 2n whenever n is a positive integer using mathematical induction. BASIS STEP: By the inductive hypothesis, 3 | (k³ + 2k), and certainly 3 | 3(k² + 1). As the sum of two multiples of 3 is again divisible by 3, 3 | ((k+1)3 + 2(k + 1)). INDUCTIVE STEP: 31 (03+2 0), i.e., 3 | 0, so the basis step is true. (k+1)3 + 2(k+1) = (k³ + 3k² + 1) + (2k + 2) = (k³ + 2k) + 3(k² + 1) Suppose that 3 | (k³ + 2k). (k+1)3 + 2(k+1) = (k³ + 3k² + 3k+ 1) + (2k + 2) = (k³ +2k)+3(k²+k+1) By the inductive hypothesis, 3 | (k³ + 2k), and certainly 3 | 3(k² + k + 1). 3|(13+2 1), i.e., 3 | 3, so the basis step is true. Reset
Click and drag the steps to their corresponding step numbers to prove that the given pair of functions are of the same order. (Note: Consider to prove the result, first prove f(x) = O(g(x)) and then prove g(x) = O(f(x)). f(x) = 2x²+x-7 and g(x) = x² Step 1 Hence, f(x) = O(g(x)) and g(x) = O(f(x)). Step 2 For large x, 2x² + x 7 ≥ 3x². Hence |f(x)|≥1g(x) for large x. For large x, x² ≤ 2x²+x-7. Hence, g(x) ≤ 1 |f(x)| for large x. Step 3 For large x, x² ≥ 2x²+x-7. Hence |f(x)|≥ 3g(x) for large x. For large x, 2x² + x 7 ≤ 3x². Hence |f(x)| ≤ 3g(x) for large x.
Transcribed Image Text:Click and drag the steps to their corresponding step numbers to prove that the given pair of functions are of the same order. (Note: Consider to prove the result, first prove f(x) = O(g(x)) and then prove g(x) = O(f(x)). f(x) = 2x²+x-7 and g(x) = x² Step 1 Hence, f(x) = O(g(x)) and g(x) = O(f(x)). Step 2 For large x, 2x² + x 7 ≥ 3x². Hence |f(x)|≥1g(x) for large x. For large x, x² ≤ 2x²+x-7. Hence, g(x) ≤ 1 |f(x)| for large x. Step 3 For large x, x² ≥ 2x²+x-7. Hence |f(x)|≥ 3g(x) for large x. For large x, 2x² + x 7 ≤ 3x². Hence |f(x)| ≤ 3g(x) for large x.
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