Question

2+5+8+11+...+(3n-1)= n(3n+1) /2​

Answer 1

Answer:

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

Answer 2

Answer:

I think 57 so I don't know exactly I think 57