Proving Divisibility by 9 using Induction

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SUMMARY

The discussion focuses on proving that the expression \(4^n + 15n - 1\) is divisible by 9 for all integers \(n > 0\) using mathematical induction. The initial step involves verifying that \(f(1) = 4^1 + 15(1) - 1\) is divisible by 9. Subsequently, the inductive step requires demonstrating that if \(f(n)\) is divisible by 9, then \(f(n+1)\) must also be divisible by 9. The problem was resolved by confirming the inductive hypothesis and applying the necessary algebraic manipulations.

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Homework Statement



prove that for all n>0 , 4^n + 15n - 1 is divisible by 9/multiple of 9

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The Attempt at a Solution


need to show: 4^(n+1) + 15(n+1) + 14/9 = a*k a and k are integers
assumption to inductive step: (4^n + 15n - 1)/9 = k --> 4^n + 15n - 1 = 9k -->
4^n+1 + 60n - 4 = 36k.. now what? I tried 4^n+1 -4 = 6(6k - 10n) but ultimately , I am stuck

please help! tank you
 
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holezch said:
need to show: 4^(n+1) + 15n + 14/9 = a*k a and k are integers
Where did you get this?

Let f(n) = 4^n + 15n - 1. First, you need to show that f(1) is divisible by 9. Then, you need to show that f(n+1) is divisible by 9 if f(n) is.
 


sorry , I had a typo. thanks for the reply but I got the answer!
 

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