FP1 - [Proof] Watch

AdeptDz
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#1
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#1
 U_1 = -1 , U_2 = 0 , U_n+2 = 6U_n+1 - 9U_n , Prove U_n = (n-2)3^{n-1}
So I done the first bit

Step 1
Prove true for n=1 and n = 2
 U_1 = (1-2)3^0 = 1

U_2 = (2-2)3^1 = 0
Both true

Step 2
Assume true for n=k and n=k+1, if true for n=k and n=k+1 then true for n=k+2

 U_k = (k-2)3^{k-1}

U_k+1 = (k-1)3^k

Prove U_k+2 = (k)3^{k+1}

U_k+2 = 6(k-1)3k-9(k-2)3^{k-1}
.....
....
....
Step 1 + Step 2 proves true for all n by induction

I don't know what to do now, may someone help me please
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B_9710
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#2
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#2
You have to show that  U_{k+2} =k(3^{k+1}) .
You are correct so far and you can get to the required result from the step you're at.
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AdeptDz
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(Original post by B_9710)
You have to show that  U_{k+2} =k(3^{k+1}) .
Yeh i know that bit thanks, i just don't know how to make  

U_k+2 = 6(k-1)3k-9(k-2)3^{k-1}
show
 U_{k+2} =k(3^{k+1})
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AdeptDz
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#4
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(Original post by B_9710)
You have to show that  U_{k+2} =k(3^{k+1}) .
You are correct so far and you can get to the required result from the step you're at.
I expanded it but i remember my teacher saying don't expand it unless you absolutely have to, and i can't tell if i have to or not like if there is another option
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NotNotBatman
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#5
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You know you need a 3^{k+1}, so look for ways to get that. What can you do with 6 and 9?
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AdeptDz
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#6
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(Original post by NotNotBatman)
You know you need a 3^{k+1}, so look for ways to get that. What can you do with 6 and 9?
take out a factor of 3?
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NotNotBatman
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#7
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#7
(Original post by AdeptDz)
divide them by 2 and 3? to get 3?
Don't divide, but factorise. Remember you can't change the value.

9 can be written as 3^x right?

And then what happens with the 3?
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AdeptDz
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#8
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#8
(Original post by NotNotBatman)
Don't divide, but factorise. Remember you can't change the value.

9 can be written as 3^x right?

And then what happens with the 3?
Yeh my bad, i editted the post after you checked it im guessing

ohh, let me try see what i get now
Thanks
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AdeptDz
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#9
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#9
So would it be
  U_k+2 = 6(k-1)3k-9(k-2)3^{k-1} 

therefore

= 

6(k-1)3k-3^2(k-2)3^{k-1}   



therefore 

= 3[2(k-1)k-3(k-2)3^{k-2}]
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NotNotBatman
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#10
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(Original post by AdeptDz)
So would it be
  U_k+2 = 6(k-1)3k-9(k-2)3^{k-1} 

therefore

= 

6(k-1)3k-3^2(k-2)3^{k-1}   



therefore 

= 3[2(k-1)k-3(k-2)3^{k-2}]
Rather than taking factorising 3 in the last step, do you notice that you'll get
in the first term 3 \times 3^k and in the second term there is a 3^2 \times 3^k remembering that a^b \times a^c =a^{b+c}

Also you've written it a bit wrong. It is
 u_{k+2} = 6(k-1)\times 3^{k} - 9(k-2) \times 3^{k-1}

remember it's 3^k, not 3k, so you can't factorise 3 like that.
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AdeptDz
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#11
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#11
(Original post by NotNotBatman)
Rather than taking factorising 3 in the last step, do you notice that you'll get
in the first term 3 \times 3^k and in the second term there is a 3^2 \times 3^k remembering that a^b \times a^c =a^{b+c}

Also you've written it a bit wrong. It is
 u_{k+2} = 6(k-1)\times 3^{k} - 9(k-2) \times 3^{k-1}

remember it's 3^k, not 3k, so you can't factorise 3 like that.
cool so:
 6(3^{k+1}-3^k) - 3^{k+2} + 18 * 3^{k-1}
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NotNotBatman
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#12
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#12
(Original post by AdeptDz)
cool so:
 6(3^{k+1}-3^k) - 3^{k+2} + 18 * 3^{k-1}
The adding power rule only applies when it's the same base, so  k \times 3^k \neq 3^{k+1}

 3^k \times 3^1 = 3^{k+1}

So you want on the first term; 2[3^1(k-1)\times 3^{k}] = 2(k-1) \times 3^1 \times 3^k, so the powers of 3 can be added. do the same with 9=3^2

But leave the brackets (k-1) and (k-2) for now.
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AdeptDz
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#13
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#13
(Original post by NotNotBatman)
The adding power rule only applies when it's the same base, so  k \times 3^k \neq 3^{k+1}

 3^k \times 3^1 = 3^{k+1}

So you want on the first term; 2[3^1(k-1)\times 3^{k}] = 2(k-1) \times 3^1 \times 3^k, so the powers of 3 can be added. do the same with 9=3^2

But leave the brackets (k-1) and (k-2) for now.
oh my bad
Would it be
 ..-3^2(k-2)3^{k-1}

so 

  2[3^1(k-1)\times 3^{k}] - 3^2(k-2)3^{k-1}
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NotNotBatman
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(Original post by AdeptDz)
oh my bad
Would it be
 ..-3^2(k-2)3^{k-1}]

so 

  2[3^1(k-1)\times 3^{k}] - 3^2(k-2)3^{k-1}
yes, then you can add the powers of 3 on each individual term.
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AdeptDz
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#15
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#15
(Original post by AdeptDz)
oh my bad
Would it be
 ..-3^2(k-2)3^{k-1}

so 

  2[3^1(k-1)\times 3^{k}] - 3^2(k-2)3^{k-1}
so


2[3^{k+1}(k-1)] - 3^{k+1}(k-2)
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NotNotBatman
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#16
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#16
(Original post by AdeptDz)
so


2[3^{k+1}(k-1)] - 3^{k+1}(k-2)
Yes, then you can factorise.
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AdeptDz
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#17
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#17
(Original post by NotNotBatman)
Yes, then you can factorise.
 (3^{k+1})(k-1)-(k+2)
Got a feeling i did that wrong as idk where to put the 2
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NotNotBatman
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#18
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(Original post by AdeptDz)
 (3^{k+1})(k-1)-(k+2)
Got a feeling i did that wrong as idk where to put the 2
Factorise the whole thing, so use a big bracket and leave the 2 on the first term, it doesn't matter which way around it's multiplied.

3^{k+1}[2(k-1)-(k+2)] when you multiply out it's the same right?
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AdeptDz
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#19
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#19
(Original post by AdeptDz)
so


2[3^{k+1}(k-1)] - 3^{k+1}(k-2)
(Original post by NotNotBatman)
Yes, then you can factorise.
1sec, let me have another go
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AdeptDz
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#20
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#20


2[3^{k+1}(k-1)] - 3^{k+1}(k-2)

2\times3^{k+1}(k-1) - 3^{k+1}(k-20)

3^{k+1}(2(k-1)-(k-2))

there
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