Use induction to prove that 2 n + 3 n is a multiple of 5 for all odd n, with n b

Question

Use induction to prove that 2n + 3n is a multiple of 5 for all odd n, with n being an element of the Natural numbers.

Step 1. Let n=1, then 2+3 = 5 which is a multiple of 5.

Step 2. Assume true for n=k, then 2k + 3k is a multiple of 5. However I do not understand how the statement only being true for odd n effects this statement. Do I assume true for 2k+1? And then show true for 2k +3? Or do I assume true for k and then show true for k+2?

If I assume true for n=k and then show true for n= k+2, I get 2k+2 + 3k+2. which equals 2k 22 + 3k32 and I know I want this to be (2k + 3k)10 which I can show is a multiple of 5. But I don’t understand how to show that.

Please walk me through the induction process.

Thank you

Explanation / Answer

We know 22k+1 + 32k+1 is divisible by 5; want to show that 22k+3 + 32k+3 is also.

Consider that 22k+3 = 4 * 22k+1 = 22k+1 + 22k+1 + 22k+1 + 22k+1

and 32k+3 = 9 * 32k+1 = 32k+1 + 32k+1 + 32k+1 + 32k+1 + 32k+1 + 32k+1 + 32k+1 + 32k+1 + 32k+1

Now the four copies of 22k+1 can pair off with the first four copies of 32k+1; those terms are divisble by 5 because of the given. But look -- that leaves 5 copies of 32k+1, or 5*32k+1, which is also divisible by 5.

QED

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