Prove that 10^{2n – 1} + 1 is divisible by 11 for all n N.

Let P(n): 10^{2n – 1} + 1 is divisible by 11

P(1): 10 + 1 = 11which is divisible by 11.

Thus P(n) is true for n = 1.

Let P(k) be true for some natural number k.

i.e. 10^{2k – 1} + 1 is divisible by 11.

Let 10^{2k – 1} + 1 = 11d

Now we prove that P(k + 1) is true whenever P(k) is true.

Now, P(k+1): 10^2k+1 + 1 = (11d – 1)100 + 1 = 11 (100d – 9),which is divisible by 11.

P(k + 1) is true.

Thus P(k +1) is true whenever P(k) is true.

By principle mathematical induction 10^{2n – 1} + 1 is divisible by 11 for all n  N.