Answer to Question #132601 in Algebra for salma

Question

prove that 4^(n)+17 is divisible by 3

Answer 1

Let's check for

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

which is divisible by 3.

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

$$f(k) = 4^k + 17 = 3 m, m\in N.$$

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

$f(k+1) = 4^{k+1} + 17 = 4^k*4 +17=4(4^k +17) - 51$ $= 4*3m - 51 = 3(4m-17)$

which is divisible by 3.

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