Prove the following by using the principle of mathematical induction for all n ∈ N: 102n – 1 + 1 is divisible by 11.

Let the given statement be $\mathrm{P}(n)$, i.e.,

$P(n): 10^{2 n-1}+1$ is divisible by $11 .$

It can be observed that $\mathrm{P}(n)$ is true for $n=1$ since $\mathrm{P}(1)=10^{2.1-1}+1=11$, which is divisible by 11 .

Let $\mathrm{P}(k)$ be true for some positive integer $k$, i.e.,

$10^{2 k-1}+1$ is divisible by 11

$\therefore 10^{2 k-1}+1=11 m$, where $m \in \mathbf{N}$

We shall now prove that P(k + 1) is true whenever P(k) is true.

Consider

$10^{2(k+1)-1}+1$

$=10^{2 h+2-1}+1$

$=10^{2 k+1}+1$

$=10^{2}\left(10^{2 k-1}+1-1\right)+1$

$=10^{2}\left(10^{2 k-1}+1\right)-10^{2}+1$

$=10^{2} .11 \mathrm{~m}-100+1 \quad[$ Using $(1)]$

$=100 \times 11 \mathrm{~m}-99$

$=11(100 m-9)$

$=11 r$, where $r=(100 m-9)$ is some natural number

Therefore, $10^{2(k+1)-1}+1$ is divisible by 11 .

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

Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.