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

Let the given statement be $P(n)$, i.e.,
$P(n) :10^{2n-1}+1$ is divisible by 11.
It can be observed that $P(n)$ is true for $n=1$ Since $P(1) =10^{2.1-1}+1$, which is divisible by 11.
Let $P(k)$ be true for some positive integer $k$, i.e.,
$10^{2k-1}+1$ is divisible by 11.
$\therefore 10^{2k-1}+1=11m$ where $m \in N$----------(1)
We shall now prove that $P(k+1)$ is true whenever $P(k)$ is true.
Consider
$10^{2(k+1)-1}+1$
$= 10^{2k+2-1}+1$
$= 10^{2k+1}+1$
$= 10^2 (10^{2k-1}+1-1)+1$
$= 10^2(10^{2k-1}+1)-10^2+1$
$=10^2.11m-100+1\qquad$[Using (1)]
$= 100 \times 11m-99$
$= 11(100m-9)$
$= 11r$, where $r=(100m-9)$ 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., $n$.
edited May 1, 2014