Question

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

Answer 1

Step $\left(1\right):$ Assume given statement
Let the given statement be $P\left(n\right)$, i.e.,
$P\left(n\right):{10}^{2n-1}+1$ is divisible by $11$.

Step $\left(2\right):$ Checking statement $P\left(n\right)$ for $n=1$
Put $n=1$ in $P\left(n\right)$, we get
$P\left(1\right):{10}^{2\cdot 1-1}+1$ is divisible by $11$
$\Rightarrow 11$ is divisible by $11$ (Which is true)
Thus $P\left(n\right)$ is true for $n=1$

Step $\left(3\right):$ $P\left(n\right)$ for $n=K$.
Put $n=K$ in $P\left(n\right)$ and consider $P\left(k\right)$ be true for some natural number $K$ i.e.,
${10}^{2K-1}+1$ is divisible by $11$
$\Rightarrow {10}^{2K-1}+1=11m$, where $m\in N\cdots \left(1\right)$

Step $\left(4\right):$ Checking statement $P\left(n\right)$ for $n=K+1$
Now, we shall prove that $P(K+1)$ is true whenever $P\left(K\right)$ is true.
Now, we have
${10}^{2(K+1)-1}+1$
$={10}^{2K+2-1}+1$
$={10}^{2K+1}+1$
$={10}^{2K-1}\cdot {10}^2+1$
$={10}^2({10}^{2K-1}+1-1)+1$
Now using $\left(1\right)$
$={10}^2(11m-1)+1$
$=11\times {10}^{2}m-{10}^2+1$
$=1100m-100+1$
$=1100m-99$
$=11(100m-9)$
$=11r\{\text{where,}r=100m-9\};r\in N$
So,${10}^{2(K+1)-1}+1$ is divisible by $11$
Thus, $P(K+1)$ is true whenever $P\left(K\right)$ is true.

Final answer :
Therefore, by the principle of mathematical induction, statement $P\left(n\right)$ is true for all $n\in N$.