FP1 - [Proof]

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     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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    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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    (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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    (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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    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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    (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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    (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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    (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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    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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    (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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    (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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    (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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    (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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    (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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    (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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    (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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    (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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    (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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    (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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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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