Question

Is ${10^{2n - 1}} + 1$ is divisible by $11$ ?

Answer 1

Here, in this question we are given a problem in which we need to check if one term is divisible by the other or not. We will use the help of the process of mathematical induction and check if we can prove the result. First we need to check if the expression is true for $n = 1$ , then we need to consider that is true for $n = k$ and then need to check if it’s true for $n = k + 1$ , if true for $n = k + 1$ we can say that the equation holds true.

Complete step by step solution:
Let $P\left( n \right):{10^{2n - 1}} + 1$ is divisible by $11$
Let us now put $n = 1$
$P\left( n \right):10 + 1 = 11$ , which is divisible by $11$
So, we get,
$\therefore P\left( n \right)$ is true for $n = 1$
Now, we assume $P\left( n \right)$ is also true for some natural number $k$ , i.e.,
$\Rightarrow {10^{2k - 1}} + 1 = 11k$ , which is divisible by $11$
$\Rightarrow {10^{2k - 1}} = 11k - 1$ -----(i)
So, we get,
$\therefore P\left( n \right)$ is true for $n = k$
$P\left( {k + 1} \right)$ must also be true whenever $P\left( n \right)$ is true. So,
$P\left( {k + 1} \right):{10^{2\left( {k + 1} \right) - 1}} + 1 \\\ P\left( {k + 1} \right):{10^{2k + 2 - 1}} + 1 \\\$
$P\left( {k + 1} \right):{10^{2k - 1}} \times {10^2} + 1$ -----(ii)
Now, from equation (i) we substitute the value of ${10^{2k - 1}}$ in equation (ii) and we get,
$P\left( {k + 1} \right):\left( {11k - 1} \right) \times {10^2} + 1 \\\ P\left( {k + 1} \right):\left( {11k - 1} \right) \times 100 + 1 \\\ P\left( {k + 1} \right):1100k - 100 + 1 \\\ P\left( {k + 1} \right):1100k - 99 \\\$
$P\left( {k + 1} \right):11\left( {100k - 9} \right)$ , which is also divisible by $11$ .
Therefore,
$\therefore P\left( n \right)$ is true for $n = k + 1$
As, $P\left( n \right)$ is true for $n = 1$ , $n = k$ and $n = k + 1$
Hence, by principal mathematical induction ${10^{2n - 1}} + 1$ is divisible by $11$ for all $n \in \mathbb{N}$ .

Note: Mathematical induction is a mathematical proof technique. It is essentially used to prove that a statement $P\left( n \right)$ holds for every natural number $n = 1,2,3,...$ , that is, the overall statement is a sequence of infinitely many cases $P\left( 0 \right),P\left( 1 \right),P\left( 2 \right),P\left( 3 \right),...$ . We generally assume that $P\left( n \right)$ is true for $n = k$ and using this we prove that $P\left( n \right)$ is true for $n = k + 1$ . Students can practise similar questions for more clarity.