How can I prove that 4^{n}+15n-1 \equiv 0 (\mod 9)?

haciendodedorcp 2022-02-28 Answered
How can I prove that 4n+15n10(mod9)?
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Expert Answer

Jonas Burt
Answered 2022-03-01 Author has 4 answers
By mathematical induction.
Clearly, the statement is true for n=1. Let it be valid for some natural number n. Is it then valid for n+1? Let us look at the difference
(4n+1+15(n+1)1)(4n+15n1)=3(4n)+15.
We see that the statement for n+1 is valid if 3(4n)+15 is divisible by 9. But is it?
Let us check it again using mathematical induction. For n=1, the number 3(4n)+15 is clearly divisible by 9. So let us assume that 3(4n)+15 is divisible by 9 for some n, and let us look at 3(4n+1)+15. We see that the difference (3(4n+1)+15)(3(4n)+15)=9(4n),
is divisible by 9, which means that 3(4n+1)+15 is divisible by 9, which means that 4n+1+15(n+1)1 is divisible by 9.
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Spandri3uv
Answered 2022-03-02 Author has 7 answers
Lets prove it mathematical induction.
Let P(n)=4n+15n1
Lets first check if it is true for n=1
P(1)=4+151=18
so 9P(1)
So it is true for n=1.
Now for induction step, we will assume it is true for n=k
P(k)=4k+15k1
And we will check for n=k+1
P(k+1)=4k+1+15(k+1)1
=4.4k+15k+151
=(1+3).4k+15k1+15
=4k+15k1+3.4k+15
=P(k)+3(4k+5)
=P(k)+3(4k1+1+5)
=P(k)+3(4k1+6)
=P(k)+3(4k1)+18
Now, as per our assumption first term P(k) is true. The term in the second bracket is divisible by 3 which is evident from the fact
(ab)(anbn)
so, (41)=3(4n1n)
3(4k1)9m(mod9).
and 918
so, P(k+1) is true.
Answer: 9=4n+15n1.
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