# Prove the following by using the principle of mathematical induction for all $n\in N$ :Q: 20            $10^{2n-1}+1$   is a divisible by $11.$

G Gautam harsolia

Let the given statement be p(n) i.e.
$p(n):10^{2n-1}+1$
For n = 1  we have
$p(1):10^{2(1)-1}+1= 10^{2-1}+1=10^1+1=11$   ,   which is divisible by 11, hence true

For  n = k  we have

$p(k):10^{2k-1}+1 \ \ \ \ \ \ \ \ \ \ \ -(i)$   ,        Let's assume that this is divisible by 11 = 11m

Now,
For  n = k + 1  we have
$p(k+1):10^{2(k+1)-1}+1$                                                                                                                                                                                                                                                                                                $=10^{2k+2-1}+1$

$=10^{2k+1}+1$
$=10^2(10^{2k-1}+1-1)+1$
$=10^2(10^{2k-1}+1)-10^2+1$
$=10^2(11m)-100+1 \ \ \ \ \ \ \ \ \ \ \ \ (using \ (i))$
$=100(11m)-99$
$=11(100m-9)$
$=11l$                          Where   $l= (100m-9)$   some natural number

Thus,  p(k+1)  is true whenever p(k) is true
Hence, by the principle of mathematical induction, statement p(n)  is divisible by 11  for all natural numbers n

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