Question

Prove by the principle of mathematical induction: 

2 + 5 + 8 + 11 + … + (3n – 1) = 1/2 n(3n + 1)

Answer 1

Suppose P (n) = 2 + 5 + 8 + 11 + … + (3n – 1) = 1/2 n(3n + 1)

Now let us check for the n = 1,

P (1): 2 = 1/2 × 1 × 4

: 2 = 2

P (n) is true for n = 1.

Now, let P (n) is true for n = k, then we have to prove that P (k + 1) is true.

P (k) = 2 + 5 + 8 + 11 + … + (3k – 1) = 1/2 k (3k + 1) … (i)

Therefore,

2 + 5 + 8 + 11 + … + (3k – 1) + (3k + 2)

Then, substituting the value of P (k) from equation (1) we get,

= 1/2 × k (3k + 1) + (3k + 2) by using equation (i)

= [3k² + k + 2 (3k + 2)] / 2

= [3k² + k + 6k + 4] / 2

= [3k² + 7k + 4] / 2

= [3k² + 4k + 3k + 4] / 2

= [3k (k + 1) + 4(k + 1)] / 2

= [(k + 1) (3k + 4)] /2

P (n) is true for n = k + 1

Thus, P (n) is true for all n ∈ N.