Question #132601
prove that 4^(n)+17 is divisible by 3
1
Expert's answer
2020-09-13T18:30:25-0400

Let's check for

n=1,f(1)=4+17=21;n = 1, f(1) = 4 + 17 = 21;

which is divisible by 3.

  1. Let's assume that f(n)=f(n) = 4n+174^n + 17 is divisible by 3 for k, kN:k\in N:
f(k)=4k+17=3m,mN.f(k) = 4^k + 17 = 3 m, m\in N.

2. Let's prove that f(k+1)f(k+1) is divisible by 3:

f(k+1)=4k+1+17=4k4+17=4(4k+17)51f(k+1) = 4^{k+1} + 17 = 4^k*4 +17=4(4^k +17) - 51 =43m51=3(4m17)= 4*3m - 51 = 3(4m-17)

which is divisible by 3.

Thus f(k+1)f(k+1) is divisible by 3 if f(k)f(k) is divisible by 3. Hence, by the mathematical induction f(n)=4n+17f(n)=4^n + 17 is divisible by 3 for nN.n\in N.



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