Prove that the following is true for all $n \in N$ using the principle of mathematical induction:
$10^{2n-1} + 1$ is divisible by $11$.

(N/A) Let the given statement be $P(n)$,i.e.,
$P(n): 10^{2n-1} + 1$ is divisible by $11$.
Step $1$: Check for $n = 1$.
$P(1) = 10^{2(1)-1} + 1 = 10^1 + 1 = 11$,which is divisible by $11$.
Thus,$P(1)$ is true.
Step $2$: Assume $P(k)$ is true for some positive integer $k$.
$10^{2k-1} + 1 = 11m$,where $m \in N$ --- $(i)$
Step $3$: Prove $P(k+1)$ is true.
Consider $10^{2(k+1)-1} + 1 = 10^{2k+2-1} + 1 = 10^{2k+1} + 1$.
$= 10^2 \cdot 10^{2k-1} + 1$
$= 100 \cdot (11m - 1) + 1$ [From $(i)$,$10^{2k-1} = 11m - 1$]
$= 1100m - 100 + 1$
$= 1100m - 99$
$= 11(100m - 9)$.
Since $11(100m - 9)$ is a multiple of $11$,$P(k+1)$ is true whenever $P(k)$ is true.
Hence,by the principle of mathematical induction,$P(n)$ is true for all $n \in N$.