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Q.
Use principle of mathematical induction to find the value of $k$, where $(10^{2n - 1} + 1)$ is divisible by $k$.
Principle of Mathematical Induction
Solution:
Let $P(n)$ be the statement given by $P(n) : 10^{2n - 1} + 1$ is divisible by $11$ For $n = 1, P(1) : 10^{(2 × 1) - 1} + 1 = 11,$ which is divisible by $11.$
So, $P(1)$ is true.
Let $P(k)$ be true, i.e. $10^{2k - 1} + 1$ is divisible by $11$
$\Rightarrow 10^{2k-1}+1=11\,\lambda$, for some $\lambda \in N \,...\left(i\right)$
We shall now show that $P(k + 1)$ is true. For this, we have to show that $10^{2(k + 1) - 1} + 1$ is divisible by $11$.
Now, $10^{2\left(k+1\right)-1}+1=10^{2k-1}\,10^{2}+1$
$=\left(11\lambda-1\right)100+1$ [Using (i)]
$=1100\lambda-99=11\left(100\lambda-9\right)=11\,\mu.$
where $\mu=100\lambda-9\in N$
$\Rightarrow 10^{2\left(k+1\right)-1}+1$ is divisible by $11$
$\Rightarrow P\left(k+1\right)$ is true.
Thus, $P\left(k + 1\right)$ is true, whenever $P\left(k\right)$ is true.
Hence, by the principle of mathematical induction, $P\left(k\right)$ is true for all $n \in N,$ i.e. $10^{2n-1}+1$ is divisible by $11$ for all $n\in N.$