Question

Use principle of mathematical induction to find the value of $k$, where $(10^{2n-1}+1)$ is divisible by $k$.

• A. $11$
• B. $12$
• C. $13$
• D. $9$

Answer 1

Answer: (A)

Let $P(n)$ be the statement given by $P(n):10^{2n-1}+1$ is divisible by $11$ For $n=1,P(1):10^{(2\times 1)-1}+1=11,$ 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 $n\in N.$