For what integers n is 3^(2n+1) + 2^(n+2) divisible by 7?
Can anybody help me?
But it's mentioned integers in the question...So 0 also must be included??
My Proof...
first the base case.
n = 0
3^1 +2^2 = 3+4 = 7
now the more general case.
let k = n. assume 3^(2*k+1) +2^(k+2) is divible by 7.
prove 3^(2*(k+1)+1) +2^((k+1)+2) is divisable by 7.
3^(2*k +3) +2^(k+3)
3^2 *3^(2*k +1) +2* 2^(k+2)
9* 3^(2*k +1) +2* 2^(k+2).
so, based on our assumption, we know 3^(2*k+1) +2^(k+2) is divisable by 7.
we also know based on our formula, that the power for 3 will always increase by 2, and the power for 2 will always increase by 1, for subequent k's. therefore 9* 3^(2*k+1) +2* 2^(k+1) will also be divisble by 7
if one is to use mathematical induction, one must know that :
set of natural numbers is subset of set with "features" :
1° (1) is natural number,
2° any natural number (n) have his follower n'=n+1 which is also natural number
3° (1) is not follower of any naturan number
4° if m'=n' than m=n, meaning every natural number is follower of one natural number (and onley one)
5° if and if in M apply properties 1° and 2° than M=N
axiom 5 is known as the principle of induction
and so on...
Edit : sorry for this... i was going somewhere else with this.
anyway can't do you harm to know this ( because this is sub forum "university math" so i thought that you would, and should know this )
Yes you should include 0 because the question asks about integers. This poses no problem with respect to using induction.
I don't see how you justified your conclusion at the end. I would rewrite
9* 3^(2*k +1) +2* 2^(k+2)
as
7 * 3^(2k + 1) + 2 * [3^(2k + 1) + 2^(k + 2)]
See?