prove this: n is a natural number 19|(5.2³ⁿ/2²+3³ⁿ/3)
how?
how?
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Prove by mathematical induction
Check for n = 1
5•2³ⁿ/2² + 3³ⁿ/3
= 5 • 2³/2² + 3³/3
= 10 + 9
= 19
Statement is true for n = 1
Assume statement is true for n = k ----> 19|(5•2³ᵏ/2²+3³ᵏ/3)
We need to show that statement is true for n = k+1
When n = k+1
5•2³ⁿ/2² + 3³ⁿ/3
= 5•2³⁽ᵏ⁺¹⁾/2² + 3³⁽ᵏ⁺¹⁾/3
= 5•2⁽³ᵏ⁺³⁾/2² + 3⁽³ᵏ⁺³⁾/3
= 2³ • 5•2³ᵏ/2² + 3³ • 3³ᵏ/3
= 8 • 5•2³ᵏ/2² + 27 • 3³ᵏ/3
= 8 • 5•2³ᵏ/2² + 8 • 3³ᵏ/3 + 19 • 3³ᵏ/3
= 8 (5•2³ᵏ/2² + 3³ᵏ/3) + 19 • 3⁽³ᵏ⁻¹⁾
This is divisible by 19 because 19|(5•2³ᵏ/2² + 3³ᵏ/3) and 19|19 • 3⁽³ᵏ⁻¹⁾
Therefore, by mathematical induction, statement is true for all integers n > 0
Check for n = 1
5•2³ⁿ/2² + 3³ⁿ/3
= 5 • 2³/2² + 3³/3
= 10 + 9
= 19
Statement is true for n = 1
Assume statement is true for n = k ----> 19|(5•2³ᵏ/2²+3³ᵏ/3)
We need to show that statement is true for n = k+1
When n = k+1
5•2³ⁿ/2² + 3³ⁿ/3
= 5•2³⁽ᵏ⁺¹⁾/2² + 3³⁽ᵏ⁺¹⁾/3
= 5•2⁽³ᵏ⁺³⁾/2² + 3⁽³ᵏ⁺³⁾/3
= 2³ • 5•2³ᵏ/2² + 3³ • 3³ᵏ/3
= 8 • 5•2³ᵏ/2² + 27 • 3³ᵏ/3
= 8 • 5•2³ᵏ/2² + 8 • 3³ᵏ/3 + 19 • 3³ᵏ/3
= 8 (5•2³ᵏ/2² + 3³ᵏ/3) + 19 • 3⁽³ᵏ⁻¹⁾
This is divisible by 19 because 19|(5•2³ᵏ/2² + 3³ᵏ/3) and 19|19 • 3⁽³ᵏ⁻¹⁾
Therefore, by mathematical induction, statement is true for all integers n > 0