Use the principal of mathematical induction to prove the following statement true for all n∈N.

Question

Use the principle of mathematical induction to prove the following statement true for all n∈N.

x + 4x + 7x + ... + (3n – 2)x = \(\frac{1}{2}\)n(3n – 1)x.

Answer 1

Let T(n) be the statement: 

x + 4x + 7x + ... + (3n – 2)x = \(\frac{1}{2}\)n(3n – 1)x 

Basic Step: For n = 1, 

x = \(\frac{1}{2}\) × 1 × (3 × 1 – 1) × x ⇒ x = x ⇒ T(1) is true. 

Induction Step: Assume T(k) holds for a natural number k, i.e., 

x + 4x + 7x + ... + (3k – 2)x = \(\frac{1}{2}\)k(3k – 1)x 

Now to show that T(k + 1) holds, add the (k + 1)^th term = [3(k + 1) – 2]x = (3k + 1)x to both the sides of T(k), i.e., 

x + 4x + 7x + ... + (3k – 2)x + (3k + 1)x = \(\frac{1}{2}\)k(3k – 1)x + (3k + 1)x 

= \(\frac{1}{2}\)[k(3k – 1)x + 2(3k + 1)x] = \(\frac{1}{2}\) [(3k² + 5k + 2) x] = \(\frac{1}{2}\)(k + 1) (3k + 2) x 

= \(\frac{1}{2}\)(k + 1) [3 (k + 1) – 1] x 

⇒ T (k + 1) is true, whenever T(k) is true.