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

Asked by Pragya Singh | 1 year ago |  130

##### Solution :-

We can write the given statement as

P (n): 102n – 1 + 1 is divisible by 11

If n = 1 we get

P (1) = 102.1 – 1 + 1 = 11, which is divisible by 11

Which is true.

Consider P (k) be true for some positive integer k

102k – 1 + 1 is divisible by 11

102k – 1 + 1 = 11m, where m ∈ N …… (1)

Now let us prove that P (k + 1) is true.

Here

10 2 (k + 1) – 1 + 1

We can write it as

= 10 2k + 2 – 1 + 1

= 10 2k + 1 + 1

By addition and subtraction of 1

= 10 2 (102k-1 + 1 – 1) + 1

We get

= 10 2 (102k-1 + 1) – 102 + 1

Using equation 1 we get

= 102. 11m – 100 + 1

= 100 × 11m – 99

Taking out the common terms

= 11 (100m – 9)

= 11 r, where r = (100m – 9) is some natural number

10 2(k + 1) – 1 + 1 is divisible by 11

P (k + 1) is true whenever P (k) is true.

Therefore, by the principle of mathematical induction, statement P (n) is true for all natural numbers i.e. n.

Answered by Abhisek | 1 year ago

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