Question

Prove ${10}^{2n-1}+1\text{is divisible by 11.}$

Answer 1

$\text{Let P (n)}={10}^{2n-1}+1\text{is divisible by 11}\text{For}n=1P\left(1\right)={10}^{2\times 1-1}+1\text{is divisible by 11}\Rightarrow 11\text{is divisible by}11\therefore P\left(1\right)\text{is true}$
Let P(n) be true for n = k
$\therefore P\left(k\right)={10}^{2k-1}+1\text{is divisible by 11}\Rightarrow {10}^{2k-1}+1=11\lambda \Rightarrow {10}^{2k-1}=11\lambda -1\text{For n = k + 1}P(k+1)={10}^{2(k+1)-1}+1\text{is divisible by 11}\Rightarrow {10}^{2k+1}+1\text{is divisible by 11}\text{Now}{10}^{2k+1}+1={10}^{2k-1}.{10}^2+1=(11\lambda -1).{10}^2+1=1100\lambda -100+1=1100\lambda -99=11(100\lambda -9)\Rightarrow {10}^{2k+1}+1\text{is divisible by 11}\therefore P(k+1)\text{is true}$
thus P (k) is true $\Rightarrow$ P (k + 1) is true
Hence by principle of mathematical induction,