Let the given statement be P(n), i.e.,
P (n) : 102n−1 + 1 is divisible by 11.
First we prove that the statement is true for n = 1,
P(1) : 102–1 + 1 = 10 + 1 = 11 is divisible by 11.
Assume P(k) is true for some positive integer k, i.e., 102k−1 + 1 is divisible by 11 ,
i.e., 102k−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 102(k+1)–1 + 1 is divisible by 11.
Let us consider, 102(k+1)–1 + 1 = 102k+2–1 + 1
= 102k–1102 + 1 = (11l – 1) 102 + 1 [From (i)]
= 1100l – 100 + 1 = 1100l – 99 = 11(100l – 9)
Thus 102(k+1)–1 + 1 is divisible by 11.
∴ P(k + 1) is true, whenever P(k) is true.