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## Main Question or Discussion Point

I need to prove by mathematical induction that all positive numbers of the form 5^n-4n+15 are divisible by 16 where n is a natural number(1,2,3,4,5...).

So far

P(1) = 5^1-4*1+15 = 16 true for P(1)

AssumeP(k) is true.

5^k-4k + 15 is disible by 16.

Now for P(k+1)

P(k+1)=5^(k+1) -4(k+1)+ 15 I don't know how to prove it is divisible by 16?

Normally with these the P(k+1) usually equals a sum of 2 numbers that collecting like terms,etc. comes to the same style as the original equation except with k+1 where n is. Thanks for your time.

So far

P(1) = 5^1-4*1+15 = 16 true for P(1)

AssumeP(k) is true.

5^k-4k + 15 is disible by 16.

Now for P(k+1)

P(k+1)=5^(k+1) -4(k+1)+ 15 I don't know how to prove it is divisible by 16?

Normally with these the P(k+1) usually equals a sum of 2 numbers that collecting like terms,etc. comes to the same style as the original equation except with k+1 where n is. Thanks for your time.