How to simplify a polynomial divided by a natural number?

errang
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Hey,

       I had a question about simplifying a polynomial divided by a natural number... I start off with

((9^n) + 24n + 15)/48

And I have to show that this works for n+1 too... So I have...

((9^(n+1)) + 24n + 24 + 15)/48

Its the 9^(n+1) that's killing me... I've tried everything I could think of, but I couldn't get rid of it...

Appreciate any tips/pointers on this.
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9^(n+1)  =  9*9^n

Author

Commented:
Yea, got that far...

I'm trying to reduce the n+ 1 polynomial to the n polynomial + some stuff.
do you need an n+1 after the 24 too?
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BTW - ((9^n) + 24n + 15) is not a polynomial    http://en.wikipedia.org/wiki/PolynomialThe 9^n term is an exponential.    http://www.purplemath.com/modules/expofcns.htm
And I have to show that this works for n+1 too (my emphasis)
 
what is the 'this' that you are refering to?
 
AW

Author

Commented:
I am basically trying to reduce ((9^(n+1)) + 24n + 24 + 15)/48 to ((9^n) + 24n + 15)/48 + some numbers.
"I am basically trying to reduce ((9^(n+1)) + 24n + 24 + 15)/48 to ((9^n) + 24n + 15)/48 + some numbers"
can't be done, since the term 9^(n+1) = 9*9^n  and that can't be shown as 9^n plus some numbers.
 ((9^(n+1)) + 24(n+1) + 15)/48  = (9^n + 24n + 15)/48 + (8*9^n + 24)/48
AW
 
You start off with:  ((9^n) + 24n + 15)/48

Where did you get this, what does it mean, and what is it for?

This isn't a very complicated expression.  You could evaluate using Excel for n = 1 ...

I've done that.  It is interesting that all the answers are integers.  Is that what you are talking about?

Also, once n is larger than 4 or so, the exponential term dominates.  So each term is approx 9x its predecessor.
1	1.00
2	3.00
3	17.00
4	139.00
5	1233.00
6	11075.00
7	99649.00
8	896811.00
9	8071265.00
10	72641347.00
11	653772081.00
12	5883948683.00
13	52955538097.00
14	476599842819.00
15	4289398585313.00
16	38604587267755.00
17	347441285409729.00

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Author

Commented:
I'm trying to show that ((9^n) + 24n + 15) divides by 48 for all n, I know that I could just go dividing it in excel, but I need to come up with a general expression.
I assume you wish to try things out rather than getting a complete solution from us.
First define your function
    f(n) = 9^n + 24n + 15                              (1)
and assume that there exists an integer k such that for some particular integer n, then
    f(n) =  48 k                                              (2)
Then, as you did in your OP, compute
    f(n+1) = 9^(n+1) + 24(n+1) + 15
You want to show that if (1) is true, then
    f(n+1) = 48 k2 for some integer k2            (3)

What you are trying to show is that if (2) is true, then (3) is also true. This is proof by induction.
    http://en.wikipedia.org/wiki/Mathematical_induction
The base case is f(1) = 48

You have already computed
    f(n+1) = 9^(n+1) + 24(n+1)  + 15

Tip #1:
Now, using d-glitch formula in first post: 9^(n+1)  =  9*9^n
    f(n+1) = 9*9^n + 24(n+1)  + 15             (4)

Tip #2:
Looking at (1) and (4), you can see a common expression, 9^n
Solve for 9^n in (1)

Tip #3:
When you do a few more steps, you will find a number of terms each having a factor of 48 in them. You can factor this 48 out and come up with the final desired result.
"I'm trying to show that ((9^n) + 24n + 15) divides by 48 for all n,
Of course it does.
But do you want the result to be an integer (or something else definite)?
"((9^n) + 24n + 15)/48

And I have to show that this works for n+1 too... So I have...

((9^(n+1)) + 24n + 24 + 15)/48"

--
If you replace the first n by n+1 you must replace the second n by n+1 too.

Author

Commented:
>>But do you want the result to be an integer (or something else definite)?

yea, I need the answer to be an Integer, I'm trying out the tips phoffric mentioned right now...

one thing though, I do realize 9^(n+1) = 9^n * 9.

I haven't tried  f(n+1) = 48 k2 for some integer k2 yet tho.
To emphasize agreement with your OP:
   f(n+1) = 9^(n+1) + 24(n+1)  + 15
             = 9^(n+1) + 24n + 24 +15
You want to show that there exists an Integer, k2, such that
   f(n+1) = 48 k2

Author

Commented:
>>((9^(n+1)) + 24n + 24 + 15)/48"

--
If you replace the first n by n+1 you must replace the second n by n+1 too.

yes, I know that, as I wrote in my original question:

>>And I have to show that this works for n+1 too... So I have...

((9^(n+1)) + 24n + 24 + 15)/48
You know that if an integer A is divisible by 48, then you can write
      A/48 = m, an integer (i.e., no remainder)
So, obviously A = 48 m

If you work with the f(n) = 9^n + 24n + 15, and not divide by 48, you may find it easier (although either way should work).
ultimately, it comes down to demonstrating that 8*9^n + 24 is always divisible by 48
8*9^n + 24 = 8*(9^n +3) so, is 9^n+3 always divisble by 6
that reduces to 8*9^n being divisible by 6 which is true (8*9^n = 4*2*(3^(2n))= 4*6*3^(2n-1) which IS divible by 6)
That proves the original hypothesis: (9^n+24n+15) is always divisible by 48
 
AW

Author

Commented:
>>ultimately, it comes down to demonstrating that 8*9^n + 24 is always divisible by 48

Sorry... bit confused how you got 8*9^n + 24
If you solve for 9^n using (1) and (2), and then plug that result into (4), then the 9^n factor disappears.
from my earlier post:
((9^(n+1)) + 24(n+1) + 15)/48  = (9^n + 24n + 15)/48 + (8*9^n + 24)/48
 
 so if (9^n + 24n + 15) is divisible by 48, and (8*9^n + 24) is also divisible by 48, then it follows that ((9^(n+1)) + 24(n+1) + 15) will be  divisible by 48, which proves the original assertion.
 
AW
Multiple ways to prove your problem.

To see the 8 * 9^n, you can compute f(n+1) - f(n) and watch most of terms cancel out.

>> (8*9^n + 24) is also divisible by 48
This shows you that it would be useful for you to work out both approaches if you want to obtain more in-depth learning. Each approach has nice things to learn.

See if you can prove (without induction) that (8*9^n + 24) is also divisible by 48.


phoffric>> why do you say "See if you can prove (without induction) that (8*9^n + 24) is also divisible by 48." (my emphasis)
what is the problem with proving that WITH induction (which I demonstrated above is quite straight forward) ?
AW

Another way to approach the problem is to separately show that the polynomial is always divisible by both 3 and 16.  

If you are familiar with modular arithmetic:

    9^1 =    9  ==>  9 mod 48          24*1 = 24 ==> 24 mod 48
    9^2 =   81  ==> 33 mod 48          24*2 = 48 ==>  0 mod 48
    9^3 =  729  ==>  9 mod 48          24*3 = 72 ==> 24 mod 48
    9^4 = 6561  ==> 33 mod 48          24*4 = 96 ==>  0 mod 48

This pattern continues ...

For n =  odd  ==>   9 + 24 + 15 = 48
For n = even  ==>  33 +  0 + 15 = 48

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Author

Commented:
Thanks
>> phoffric>> why do you say "See if you can prove (without induction) that (8*9^n + 24) is also divisible by 48." (my emphasis) what is the problem with proving that WITH induction (which I demonstrated above is quite straight forward) ?

We have been showing multiple families of solutions. I offered another one since it is not obvious after looking at:
(9^n + 3) for a second or two that this expression is divisible by 2.

If the author goes through every approach, then there will be a nice bag of tricks for the future.

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