n3 – n is divisible by 6, for each natural number n ≥ 2

1.	n3 – n is divisible by 6, for each natural number n ≥ 2
Answer» 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.

Answer»

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.

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