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.
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.
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
In order to show that 3^(2n) - 1 is divisible by 8 for all natural numbers n, we can use mathematical induction. Here's how to fill in the missing parts:
P(1) is the statement that 3^(2*1) - 1 = 3^2 - 1 = 9 - 1 = 8 is divisible by 8, which is true.
Assume that P(k) is true. Thus, our induction hypothesis is 3^(2k) - 1 is divisible by 8.
We want to use this to show that P(k + 1) is true. Now,
3^(2(k+1)) - 1 = 3^(2k+2) - 1
= 3^2 * 3^2k - 1
= 9 * (3^(2k) - 1) + 8
Here, we can apply the induction hypothesis. We assumed that 3^(2k) - 1 is divisible by 8, so let's say 3^(2k) - 1 = 8a, where a is some natural number.
Substituting this into the equation,
9 * (3^(2k) - 1) + 8 = 9 * (8a) + 8 = 72a + 8
Now, we can rewrite 72a + 8 as 8 * (9a + 1), which is divisible by 8.
This final result is divisible by 8, since 9a + 1 is divisible by 8 (since 9a + 1 is an integer). 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.
To prove that 3^2n - 1 is divisible by 8 for all natural numbers n, we can use mathematical induction.
First, let's verify the base case, which is P(1). Here, we need to show that 3^2(1) - 1 is divisible by 8. Simplifying this expression gives us 9 - 1 = 8, which is indeed divisible by 8. Hence, the base case is true.
Now, let's assume that P(k) is true, where k is any natural number. This means that 3^2k - 1 is divisible by 8.
Next, we need to show that P(k + 1) is true using the induction hypothesis.
We start with the expression 3^2(k + 1) - 1 and try to express it in terms of 3^2k - 1.
3^2(k + 1) - 1 = 3^2k * 3^2 - 1
Since 3^2 = 9, we can substitute this value:
3^2(k + 1) - 1 = 3^2k * 9 - 1
Expanding this expression gives us:
3^2k * 9 - 1 = 9 * 3^2k - 1
Notice that 9 * 3^2k is divisible by 8 because it is a multiple of 8 (9 * 3^2k = 8 * 3^2k + 1).
Hence, 9 * 3^2k - 1 is also divisible by 8 since 1 is divisible by 8.
By completing the induction step, we have shown that if P(k) is true, then P(k + 1) is also true.
By the principle of mathematical induction, we can conclude that P(n) is true for all natural numbers n. Therefore, 3^2n - 1 is divisible by 8 for all natural numbers n.