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IV. Prove the following using mathematical induction. 1. 5" – 1 is divisible by 4, Vn € W. 2. 8" – 3" is divisible by 5, Vn E W. 3. x2n – y2n has a factor x +y, Vn E N. - 4. n? < n!, for any integer n > 4 5. (a +1)" >1+ na, where a > -1, Vn E N 1 6. 3+32 + 3³ +...+ 3ª = (3n+1 – 3), Vn E N. 2 | 7. 12 + 32 + 52 + ...+ (2n – 1)? = n(2n – 1)(2n +1) Vn E N. | 3 (2n – 1)3n+1 + 3 8. 1.3+2.32 + 3 · 33 + ... + n · 3ª = Vn E N. 4 n°(n+1)² 9. 13 + 23 + 33 + ...+ n³ Vn E N. %3D 4 1 10. 1.2 1 1 n +. 3.4 ... 2-3 n. (n+1) n+1'

IV. Prove the following using mathematical induction. 1. 5" – 1 is divisible by 4, Vn € W. 2. 8" – 3" is divisible by 5, Vn E W. 3. x2n – y2n has a factor x +y, Vn E N. - 4. n? < n!, for any integer n > 4 5. (a +1)" >1+ na, where a > -1, Vn E N 1 6. 3+32 + 3³ +...+ 3ª = (3n+1 – 3), Vn E N. 2 | 7. 12 + 32 + 52 + ...+ (2n – 1)? = n(2n – 1)(2n +1) Vn E N. | 3 (2n – 1)3n+1 + 3 8. 1.3+2.32 + 3 · 33 + ... + n · 3ª = Vn E N. 4 n°(n+1)² 9. 13 + 23 + 33 + ...+ n³ Vn E N. %3D 4 1 10. 1.2 1 1 n +. 3.4 ... 2-3 n. (n+1) n+1'

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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IV. Prove the following using mathematical induction. 1. 5" – 1 is divisible by 4, Vn € W. 2. 8" – 3" is divisible by 5, Vn E W. 3. x2n – y2n has a factor x +y, Vn E N. - 4. n? < n!, for any integer n > 4 5. (a + 1)" >1+ na, where a > -1, Vn E N 1 6. 3+32 + 3³ +...+ 3ª = (3n+1 – 3), Vn E N. 2 | 7. 12 + 32 + 52 + ...+ (2n – 1)? = n(2n – 1)(2n +1) Vn E N. | 3 (2n – 1)3n+1 + 3 8. 1.3+2.32 + 3 · 33 + ... + n · 3ª = Vn E N. 4 n°(n+1)² 9. 13 + 23 + 33 + ...+ n³ Vn E N. %3D 4 1 10. 1.2 1 1 n +. 3.4 ... 2-3 n. (n+1) n+1'
Transcribed Image Text:IV. Prove the following using mathematical induction. 1. 5" – 1 is divisible by 4, Vn € W. 2. 8" – 3" is divisible by 5, Vn E W. 3. x2n – y2n has a factor x +y, Vn E N. - 4. n? < n!, for any integer n > 4 5. (a + 1)" >1+ na, where a > -1, Vn E N 1 6. 3+32 + 3³ +...+ 3ª = (3n+1 – 3), Vn E N. 2 | 7. 12 + 32 + 52 + ...+ (2n – 1)? = n(2n – 1)(2n +1) Vn E N. | 3 (2n – 1)3n+1 + 3 8. 1.3+2.32 + 3 · 33 + ... + n · 3ª = Vn E N. 4 n°(n+1)² 9. 13 + 23 + 33 + ...+ n³ Vn E N. %3D 4 1 10. 1.2 1 1 n +. 3.4 ... 2-3 n. (n+1) n+1'
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