Step (1): Assume given statement
Let the given statement be P(n), i.e.
P(n):7n−3n is divisible by 4.
Step (2): Checking statement P(n) for n=1
Put n=1 in P(n), we get
=71−31=7−3=4 which is divisible by 4.
Thus, P(n) is true for n=1.
Step (3): P(n) for n=K
Put n=K in P(n), and assume that P(K) is true for some
natural number K i.e.
P(K)=7K−3K is divisible by 4.
We can write
7K−3K=4d, where d∈N ⋯(1)
Step (4): Checking statement P(n) for n=K+1
Now, we shall prove that P(K+1) is true whenever P(K) is true.We have
7K+1−3K+1
=7K+1−7⋅3K+7⋅3K−3K+1
=7(7K−3K)+3K(7−3)
=7(4d)+4⋅3K
=4(7d+3K)
From the last line, we can see that
7K+1−3K+1 is divisible by 4.
Thus, P(K+1) is true when P(K) is true.
Final Answer:
Therefore, by principle of Mathematical Induction, the statement P(n) is true for every positive integer n.