Let's check for
n=1,f(1)=4+17=21;-
which is divisible by 3.
- Let's assume that f(n)= 4n+17 is divisible by 3 for k, k∈N:
f(k)=4k+17=3m,m∈N. 2. Let's prove that f(k+1) is divisible by 3:
f(k+1)=4k+1+17=4k∗4+17=4(4k+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)=4n+17 is divisible by 3 for n∈N.
Comments