Question -
Answer -
Let P (n) = 2 + 5 + 8 + 11 + … + (3n – 1) = 1/2 n (3n + 1)
Let us check for n = 1,
P (1): 2 = 1/2 × 1 × 4
: 2 = 2
P (n) is true for n = 1.
Now, let us check for P (n) is true for n = k, and have toprove that P (k + 1) is true.
P (k) = 2 + 5 + 8 + 11 + … + (3k – 1) = 1/2 k (3k + 1) … (i)
So,
2 + 5 + 8 + 11 + … + (3k – 1) + (3k + 2)
Now, substituting the value of P (k) we get,
= 1/2 × k (3k + 1) + (3k + 2) by using equation (i)
= [3k2 + k + 2 (3k + 2)] / 2
= [3k2 + k + 6k + 2] / 2
= [3k2 + 7k + 2] / 2
= [3k2 + 4k + 3k + 2] / 2
= [3k (k+1) + 4(k+1)] / 2
= [(k+1) (3k+4)] /2
P (n) is true for n = k + 1
Hence, P (n) is true for all n ∈ N.