Let the given statement be P(n), i.e., P (n) : $10^{2n-1}$ + 1 is divisible by 11. First we prove that the statement is true for n = 1, P(1) : $10^{2 – 1}$ + 1 = 10 + 1 = 11 is divisible by 11. Assume P(k) is true for some positive integer k, i.e., $10^{2k-1}$ + 1 is divisible by 11 , i.e., $10^{2k-1}$ + 1 = 11 l , where l ∊ N ... (i) Now we shall prove that P(k + 1) is true. For this we have to prove that $10^{2(k + 1)–1}$ + 1 is divisible by 11. Let us consider, $10^{2(k + 1) – 1}$ + 1 = $10^{2k + 2 – 1}$ + 1 = $10^{2k – 1} 10^2$ + 1 = (11l – 1) $10^2$ + 1 [From (i)] = 1100l – 100 + 1 = 1100l – 99 = 11(100l – 9) Thus $10^{2(k + 1) – 1}$ + 1 is divisible by 11. ∴ P(k + 1) is true, whenever P(k) is true.