Answer :

Let P(n) = 3

^{2n+2}– 8n – 9

P(1) is divisible by 8

Let us assume P(k) is divisible by 8.

P(k) = 3

^{2k+2}– 8k – 9 = 8M………….. (A)

To prove P(k+1) is divisible by 8 using the result of (A)

P(k+1) = 3

^{2(k+1)+2}– 8(k+1) – 9

= 3

^{2k+2+2}– 8k – 8 –9

= 3

^{2k+2}. 3

^{2}– 8k- 9 –8

= (8M + 8K + 9) 9 – 8k – 9 – 8

= 72 M + 72 K + 81 – 8K – 9 – 8

= 72 M + 64 k +64

= 8(9M + 8K + 8)

P( K + 1) is divisible by 8.

hence the result.

P(K+1) is true.

By the Principle of mathematical induction, P(n) is true for all values of nwhere n N

Hence proved

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