Question

so this is a fill in on a worksheet and I am having difficulty as the ones I inserted are incorrect can anybody help me how to do it all, sorry it's a long problem.
Show that 3^2n − 1 is divisible by 8 for all natural numbers n.

Let P(n) denote the statement that 3^2n − 1
is divisible by 8.

P(1) is the statement that
3^??? − 1 = ???? is divisible by 8, which is true.

Assume that P(k) is true. Thus, our induction hypothesis is
3^??? − 1 is divisible by 8.

We want to use this to show that P(k + 1) is true. Now,
3^2( ? )−1 = ^9(3^ ?? ) -1

= 9 (3^ ?? ) - ??? +8

= 9 ( ???????) +8


This final result is divisible by 8, since ??????
is divisible by 8 by the induction hypothesis. Thus, P(k + 1) follows from P(k), and this completes the induction step. Having proven the above steps, we conclude by the Principle of Mathematical Induction that P(n) is true for all natural numbers n.

Answer 1

P: 3^(2n)-1 is a multiple of 8
P(1): 3^2-1 = 8. True

Assume P(k). Then with n=k+1, we have

3^(2(k+1))-1
= 3^(2k+2)
= 3^(2k)*9 - 1
= 3^(2k)*8 + 3^(2k) - 1
but, we know that 3^(2k)-1 is a multiple of 8, say, 8m. Then we have
=8(3^(2k))+8m
= 8(3^(2k)+m)
which is a multiple of 8.
So, P(k) ==> P(k+1)
and we are done.

Answer 2

yes for n=1 3^2 -1 = 8

3^2(n+1) - 1 = 3^2n * 3^2 - 1
so
3^2(n+1) - 1 = 3^2n * 9 -1

= 3^2n * 8 + 3^2n-1
= 8 times something + the one before