Prove that following by using the principle of mathematical induction for all $$n\in N:$$ $$3^{2n+2}-8n-9$$ is divisible by 8.



Asked by Pragya Singh | 1 year ago |  136

##### Solution :-

We can write the given statement as

P (n): 32n + 2 – 8n – 9 is divisible by 8

If n = 1 we get

P (1) = 32 × 1 + 2 – 8 × 1 – 9 = 64, which is divisible by 8

Which is true.

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

32k + 2 – 8k – 9 is divisible by 8

32k + 2 – 8k – 9 = 8m, where m ∈ N …… (1)

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

Here

2(k + 1) + 2 – 8 (k + 1) – 9

We can write it as

= 3 2k + 2 . 32 – 8k – 8 – 9

By adding and subtracting 8k and 9 we get

= 32 (32k + 2 – 8k – 9 + 8k + 9) – 8k – 17

On further simplification

= 32 (32k + 2 – 8k – 9) + 32 (8k + 9) – 8k – 17

From equation (1) we get

= 9. 8m + 9 (8k + 9) – 8k – 17

By multiplying the terms

= 9. 8m + 72k + 81 – 8k – 17

So we get

= 9. 8m + 64k + 64

By taking out the common terms

= 8 (9m + 8k + 8)

= 8r, where r = (9m + 8k + 8) is a natural number

So 3 2(k + 1) + 2 – 8 (k + 1) – 9 is divisible by 8

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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