Prove that 32n+2 - 8n - 9 is divisible by 8 using P.M.I
Answers
3^(2n+2)-8n-9 is divisible by 8 for all natural values of n.
First, consider the value of 3^(2n+2)-8n-9 for n = 1, it is 3^(2+2)-8-9 = 81 - 17 = 64
Now assume 3^(2n+2)-8n-9 is divisible by 8 for a value of n. With this assumption we test whether 3^(2n+2)-8n-9 is divisible by 8 for n = n+1
3^(2(n+1)+2)-8(n+1)-9
= 3^(2n+2+2)-8n- 8-9
= 3^(2n+2)*9 - 8n - 9 - 8
= 3^(2n+2) - 8n - 9 + 3^(2n+2)*8 - 8n - 8
= 3^(2n+2) - 8n - 9 + 8*(3^(2n+2) - n - 1)
This is clearly divisible by 8 as we have assumed 3^(2n+2) - 8n - 9 is divisible by 8 and 8*(3^(2n+2) - n - 1) has 8 as a factor.
This proves that 3^(2n+2)-8n-9 is divisible by 8 for all natural values of n
ARUV | STUDENT
Prove 3^(2n+2)-8n-9 is divisible by 8 for all natural numbers. We prove it by method of mathematical induction.
Define statement P(n):3^(2n+2)-8n-9 is divisible by 8.
P(1): 3^(2xx1+2)-8xx1-9=64/8=8
Thus P(1) is true. Let P(n) is true for n=k.
P(k):3^(2k+2)-8k-9 is divisible by 8.
3^(2k+2)-8k-9=8m (say)
We wish to prove
P(k+1) is true whenever P(k) is true.
P(k+1):3^(2(k+1)+2)-8(k+1)-9
=3^(2k+2+2)-8(k+1)-9
=3^2xx3^(2k+2)-8k-8-9
=9xx3^(2k+2)-72k+64k-9-8
=9(3^(2k+2)-8k-9)+64k+72-8
=9(8m)+64(k+1)
and 72m and 64(k+1) are divisible by 8 because 72 and 64 are divisible by 8.
Thus P(k+1) is true when P(k) is true. Thus P(n) is true for all n, natural number.
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