32n+2 - 8n - 9 is divisible by 8.

1.	32n+2 - 8n - 9 is divisible by 8.
Answer» Let P(n): 3²ⁿ⁺² - 8n - 9 is divisible by 8. For n = 1, P(1): 3^{2 + 2} – 8(2) – 9 = 64 is divisible by 8, which is true. P( 1) is true. Let us assume P(k) is true for some k ∈ N i.e., 3^2k+2 - 8k - 9 is divisible by 8. Let 3^2k+2- 8k - 9 = 8d, d ∈ N ………………(1) Consider 3^2(k+1)+2 – 8(k +1) – 9 = 3^2k+4 – 8k – 8 – 9 = 3^2k+2 - 3² - 8(k - 1) = (8d + 8k + 9)9-8k-17 using (1) = (8d + 8k)9 + 81- 8k - 17 = 8(d + k)9 + 64 - 8k = 8[9(d + k) + 8 - k] which is divisible by 8. Thus, P(k) ⇒ P(k +1) Hence, by mathematical induction, P(n) is true for all n∈N

Answer»

Let P(n): 3²ⁿ⁺² - 8n - 9 is divisible by 8.

For n = 1, P(1): 3^{2 + 2} – 8(2) – 9 = 64 is divisible by 8, which is true.

P( 1) is true.

Let us assume P(k) is true for some k ∈ N

i.e., 3^2k+2 - 8k - 9 is divisible by 8.

Let 3^2k+2- 8k - 9 = 8d, d ∈ N ………………(1)

Consider 3^2(k+1)+2 – 8(k +1) – 9 = 3^2k+4 – 8k – 8 – 9

= 3^2k+2 - 3² - 8(k - 1) = (8d + 8k + 9)9-8k-17 using (1)

= (8d + 8k)9 + 81- 8k - 17

= 8(d + k)9 + 64 - 8k = 8[9(d + k) + 8 - k]

which is divisible by 8.

Thus, P(k) ⇒ P(k +1)

Hence, by mathematical induction, P(n) is true for all n∈N

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