Let P(n) : n3 – n
Step 1 :
P(2): 23 – 2 = 6 which is divisible by 6. So it is true for P(2).
Step 2 :
P(A): k3 – k = 6λ. Let it is be true for k ≥ 2
⇒ k3 = 6λ + k … (i)
Step 3 :
P(k + 1) = (k + 1)3 – (k + 1)
= k3 + 1 + 3k2 + 3k – k – 1 = k3 – k + 3(k2 + k)
= k3 – k + 3(k2 + k) = 6λ + k – k + 3(k2 + k)
= 6λ + 3(k2 + k) [from (i)]
We know that 3(k2 + k) is divisible by 6 for every value of k ∈ N.
Hence P(k + 1) is true whenever P(k) is true.