Using the principle of induction, prove that 64 is a factor of 3^(2n+2)- 8n-9 ∀ n∈N
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Expert's answer
2022-01-31T18:08:46-0500
Let p(n)=3"^{2n+2}"−8n−9 is divisible by 64 …..(1) When put n=1,p(1)=34−8−9=64 which is divisible by 64
Let n=k and we get p(k)=3"^{2k+2}" −8k−9 is divisible by 64"3^{2k+2}-8k\u22129=64m"where m∈N …..(2)
Now we shall prove that p(k+1) is also true
p(k+1)="3^{2(k+1)+2}\u22128(k+1)\u22129" is divisible by 64.Now,
p(k+1)="3^{2(k+1)+2}\u22128(k+1)\u22129" ="3^23^{2k+2}\u22128(k+1)\u22129" = "9(3^{2k+2})\u22128(k+1)\u22129" =9(64m+8k+9)−8k−17=9(64m)+72k+81−8k−17=9(64m)+64k+64=64(9m+k+1), Which is divisibility by 64
Thus p(k+1) is true whenever p(k) is true.
Hence, by principal mathematical induction,p(x) is true for all natural number.
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