- #1
Jaroslav
- 4
- 0
I'm still learning English, had to use dictionary and translator, so I'm sorry if its unclear, i will try to explain it more if needed.
For n belonging to N when n is even and n > 3, prove that
(4^(n-3) + 5^(n-3) + 9) is divisible by 9
3. The Attempt at a Solution
[/B]
P(n) = 4^(n-3) + 5^(n-3)+9
for n = 4 :
P(4) = 4^1 + 5^1 + 9 = 18, divisible by 9
Now assume that n=k, P(k) = 4^(k-3) + 5^(k-3)+9 is div by 9, then solve it for P(k+2) = 4^(k-1) + 5^(k-1)+9
And here is the problem, i know that i should use assumption from n=k, but i really don't know how. I have problem with rearranging things to involve it...this is what i tried to use somehow, but always get stuck without any clear step forward
4^(k-3) + 5^(k-3)+9 = 4^(-2)*4^(k-1)+5^(-2)*5^(k-1)+9 = 9m
Somehow i wanted to get that (k-1) exponent from it, but i can't deal with multipliers there
Homework Statement
For n belonging to N when n is even and n > 3, prove that
(4^(n-3) + 5^(n-3) + 9) is divisible by 9
Homework Equations
3. The Attempt at a Solution
[/B]
P(n) = 4^(n-3) + 5^(n-3)+9
for n = 4 :
P(4) = 4^1 + 5^1 + 9 = 18, divisible by 9
Now assume that n=k, P(k) = 4^(k-3) + 5^(k-3)+9 is div by 9, then solve it for P(k+2) = 4^(k-1) + 5^(k-1)+9
And here is the problem, i know that i should use assumption from n=k, but i really don't know how. I have problem with rearranging things to involve it...this is what i tried to use somehow, but always get stuck without any clear step forward
4^(k-3) + 5^(k-3)+9 = 4^(-2)*4^(k-1)+5^(-2)*5^(k-1)+9 = 9m
Somehow i wanted to get that (k-1) exponent from it, but i can't deal with multipliers there