Let P(n) : n3 – 7n + 3
Step 1:
P(1) = (1)3 – 7(1) + 3
= 1 – 7 + 3 = -3 which is divisible by 3
So, it is true for P(1).
Step 2:
P(k) : k3 – 7k + 3 = 3λ. Let it be true
⇒ k3 = 3λ + 7k – 3
Step 3:
P(k + 1) = (k + 1)3 – 7(k + 1) + 3
= k3 + 1 + 3k2 + 3k – 7k – 7 + 3
= k3 + 3k2 – 4k – 3
= (3λ + 7k – 3) + 3k2 – 4k – 3 (from Step 2)
= 3k2 + 3k + 3λ – 6
= 3(k2 + k + λ – 2) which is divisible by 3.
So it is true for P(k + 1).
Hence, P(k + 1) is true whenever it is true for P(k).