prove that for any positive integer n, the value of 3^2n+2 - 8n-9 is divisible by 64
The way you typed it, 3^2n+2 - 8n-9
it isn't even true for n = 1
You must have meant:
3^(2n+2) - 8n-9
let's use induction:
step 1. is it true for n = 1 ?
3(4) - 8(1) - 9 = 64, which is clearly divisible by 64
step 2. Assume it is true for n = k
i.e. 3^(2k+2) - 8k - 9 is divisible by 64
step 3. Is 3^(2(k+1) + 2) - 8(k+1) - 9 divisible by 64
it is if the difference between the expressions in step 3 and step 2 is divisible by 64. Let's take that difference.
(concept: the difference between any two multiples of p must be divisible by p)
e.g. 72 and 192 are both multiples of 12, then 192-72 = 120, which is a multiple of 12
3^(2(k+1) + 2) - 8(k+1) - 9 - (3^(2k+2) - 8k - 9)
= 3^(2k + 4) - 8k - 8 - 9 - 3^(2k+2) + 8k + 9
= 3^(2k+4) - 3^(2k+2) - 8
= 3^(2k+2)( 3^2 - 1) - 8
= 3^(2k+2)(8) - 8
= 3^(2k)(3^2)(8) - 8
= 72(3^(2k)) - 8
= 72(3^2)^k - 8
= 72(9^k) - 8
= 8(9^(k+1)) - 8
= 8( 9^(k+1) - 1) ****
argghhh, now I have to show that 9^(k+1) - 1 is divisible by 8
step 1 , works for k=1
step 2, assum 9^(k+1) - 1 is divisible by 8
step 3, what about 9^(k+2) - 1 ???
again take the difference .....
9^(k+2) - 1 - (9^(k+1) - 1)
= 9^(k+2) - 9^(k+1)
= 9^(k+1)( 9^1 - 1)
= 9^(k+1)(8), which means 9^(k+2) - 1 is divisible by 8, Yeahhh
so in **** we have 8(multiple of 8) = a multiple of 64
To prove that the expression 3^(2n+2) - 8n - 9 is divisible by 64 for any positive integer n, we can use mathematical induction.
Step 1:
Let's start by verifying the statement for n = 1, the base case.
When n = 1, the expression becomes:
3^(2*1+2) - 8*1 - 9 = 3^4 - 8 - 9 = 81 - 8 - 9 = 64
The result is 64, which is divisible by 64. Thus, the base case holds true.
Step 2:
Assume the statement is true for some arbitrary positive integer k, where k ≥ 1.
That is, 3^(2k+2) - 8k - 9 is divisible by 64.
Step 3:
We need to show that the statement is also true for the next positive integer, k + 1.
We substitute n = k + 1 into the given expression:
3^(2(k+1)+2) - 8(k+1) - 9 = 3^(2k+4) - 8k - 8 - 9
Expanding the expression:
= 3^(2k+2) * 3^2 - 8k - 8 - 9
= (3^(2k+2) - 8k - 9) * 9 - 72
By the assumption in step 2, we know that 3^(2k+2) - 8k - 9 is divisible by 64. Let's denote it as x, where x = 64m, for some positive integer m.
Therefore, the expression becomes:
= (64m) * 9 - 72
= 576m - 72
Step 4:
To prove that the expression (576m - 72) is divisible by 64, we can rewrite it as:
= 64 * (9m - 1)
= 64 * (9m - 1)
Since 64 and (9m - 1) are both integers, we can see that (576m - 72) is divisible by 64.
Step 5:
Considering that the statement holds true for n = 1 (base case) and if it holds for k, it also holds for k + 1 (inductive step), we can conclude that the statement is proven true for all positive integers n using mathematical induction.
Therefore, for any positive integer n, the value of 3^(2n+2) - 8n - 9 is divisible by 64.