G10 Math (M2) Supplementary Exercises Chapter 2 Mathematical Induction

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G10 Mathematics (Module 2) Supplementary Exercise Chapter 2

Mathematical Induction

Section A 1.

Prove by Mathematical induction that

1 ⋅ 5 + 2 ⋅ 6 + ... + n(n + 4) =

2.

n(n + 1)(2n + 13) 6

for all positive integers n.

Prove by Mathematical induction that

2 + 3 ⋅ 2 + 4 ⋅ 2 2 + 5 ⋅ 2 3 + ... + (n + 1)2 n −1 = n ⋅ 2 n for all positive integers n. 3.

Prove by Mathematical induction that

1 ⋅ 2 2 + 2 ⋅ 3 2 + 3 ⋅ 4 2 + ... + n(n + 1) 2 =

n(n + 1)(n + 2)(3n + 5) for all positive 12

integers n. 4.

Prove by Mathematical induction that

T1 × T2 × ... × Tn =

5.

n+2 n(n + 2) , where Tn = for all positive integers n. 2(n + 1) (n + 1) 2

(a) Prove by Mathematical Induction that

1 1 1 1 n + + + ... + = for all positive integers n. 1⋅ 2 2 ⋅ 3 3 ⋅ 4 n(n + 1) n + 1 (b) Hence, find the smallest positive integers n, such that

1 1 1 1 9 + + + ... + ≥ 1⋅ 2 2 ⋅ 3 3 ⋅ 4 n(n + 1) 10

6.

Prove by mathematical induction that 7 n − 6n − 1 is divisible by 36 for all positive integers n.

7.

Prove by mathematical induction that 4 2 n+1 + 3 n + 2 is divisible by 13 for all positive integers n .

8.

Prove by mathematical induction that 4 n + 5 n is divisible by 9 for all positive odd integers n .

G10 Math (M2) Supplementary Exercises Chapter 2 Mathematical Induction

9.

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Prove by mathematical induction that n 3 + 5n is divisible by 6 for all positive integers n .

Section B 10. (a) Prove by mathematical induction that 1 ⋅ 2 2 + 2 ⋅ 3 2 + 3 ⋅ 4 2 + L + n(n + 1) 2 =

1 n(n + 1)(n + 2)(3n + 5) 12

for all positive integers n. (b) Hence evaluate 2 ⋅ 2 2 + 4 ⋅ 3 2 + 6 ⋅ 4 2 + L + 20 ⋅ 112 .

11. (a) Prove by mathematical induction that

1(3) + 2(3 2 ) + 3(33 ) + ... + n(3 n ) =

1 3 + (2n − 1)3 n +1 4

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for all positive

integers n . (b) Show that 2(3) + 3(3 2 ) + 4(33 ) + ... + 21(3 20 ) =

1 41(3 21 ) − 3 . 4

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12. (a) Prove by mathematical induction that 5 n − 1 is divisible by 4 for all positive integers n . (b) Hence prove by mathematical induction that 2(7 n ) − 3(5 n ) + 1 is divisible by 24 for all integers n ≥ 2 .

13. (a) Prove by mathematical induction that n(n + 1) is divisible by 2 for all positive integers n. (b) Hence, or otherwise, prove by mathematical induction that n 3 + 5n is divisible by 6 for all positive integers n.