Question

prove that
${10}^{2n - 1} + 1$
is divisible by 11​ by mathematical induction

Answer 1

Answer:

Step-by-step explanation:

Given Prove that   {10}^{2n - 1} + 1

is divisible by 11 by mathematical induction

We need to prove  

So let p(n) : 10^2n-1 + 1 = 11d where d belongs to N

For n = 1 it will be 11. So p(n) is true for n = 1.

Now to prove for p(k + 1) is true.

Now 10^2(k + 1) – 1 + 1

So 10^2k + 2 – 1 + 1

We can write as

10^(2k – 1) + 2 + 1

10^2k – 1 . 10^2 + 1

We have 10^2k – 1 + 1 = 11m

                10^2k – 1 = 11 m – 1

        (11m – 1). 10^2 + 1

       (11m – 1) x 100 + 1

        100 x 11 m – 99

       100 x 11 m – 9 x 11

      Or 11 (100 m – 9)

        11 x where x is some natural number.

Therefore p(k + 1) is true whenever p(k) is true.

So from principle of mathematical induction, p(n) is true for n, where n is a natural number.