Prove the statement by the Principle of Mathematical Induction :n3 �

1.	Prove the statement by the Principle of Mathematical Induction :n3 – 7n + 3 is divisible by 3, for all natural numbers n.
Answer» According to the question, P(n) = n³ – 7n + 3 is divisible by 3. So, substituting different values for n, we get, P(0) = 0³ – 7×0 + 3 = 3 which is divisible by 3. P(1) = 1³ – 7×1 + 3 = −3 which is divisible by 3. P(2) = 2³ – 7×2 + 3 = −3 which is divisible by 3. P(3) = 3³ – 7×3 + 3 = 9 which is divisible by 3. Let P(k) = k³ – 7k + 3 be divisible by 3 So, we get, ⇒ k³ – 7k + 3 = 3x. Now, we also get that, ⇒ P(k+1) = (k+1)³ – 7(k+1) + 3 = k³ + 3k² + 3k + 1 – 7k – 7 + 3 = 3x + 3(k² + k – 2) is divisible by 3. ⇒ P(k+1) is true when P(k) is true. Therefore, by Mathematical Induction, P(n) = n³ – 7n + 3 is divisible by 3, for all natural numbers n.

Prove the statement by the Principle of Mathematical Induction :n3 – 7n + 3 is divisible by 3, for all natural numbers n.

Answer»

According to the question,

P(n) = n³ – 7n + 3 is divisible by 3.

So, substituting different values for n, we get,

P(0) = 0³ – 7×0 + 3 = 3 which is divisible by 3.

P(1) = 1³ – 7×1 + 3 = −3 which is divisible by 3.

P(2) = 2³ – 7×2 + 3 = −3 which is divisible by 3.

P(3) = 3³ – 7×3 + 3 = 9 which is divisible by 3.

Let P(k) = k³ – 7k + 3 be divisible by 3

So, we get,

⇒ k³ – 7k + 3 = 3x.

Now, we also get that,

⇒ P(k+1) = (k+1)³ – 7(k+1) + 3

= k³ + 3k² + 3k + 1 – 7k – 7 + 3

= 3x + 3(k² + k – 2) is divisible by 3.

⇒ P(k+1) is true when P(k) is true.

Therefore, by Mathematical Induction,

P(n) = n³ – 7n + 3 is divisible by 3, for all natural numbers n.