(n+1)(n-1) Indices

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  1. SubAtomic's Avatar
    1 05-07-2012  20:19
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    (n+1)(n-1) Indices
    Hi, this may be sixth form, not sure.

    How does this work, cannot quite fathom it, no help in book so am left in ambiguity.

    So, I am supposed to show that \displaystyle U_{n-1}U_{n+1}-U^2_n=12^{n-1} , for \displaystyle n=1,2,3...

    \displaystyle U_n=3^n+4^n

    \displaystyle U_{n-1}U_{n+1}=(3^{n-1}+4^{n-1})(3^{n+1}+4^{n+1}) \\ \\ =3^{2n}+3^{n-1}4^{n+1}+4^{n-1}3^{n+1}+4^{2n}

    fine


    \displaystyle U_n^2=(3^n+4^n)^2=3^{2n}+2(3^n4^ n)+4^{2n}

    fine


    Now, what I don't get is how this line becomes the next

    \displaystyle 3^{n-1}4^{n+1}+4^{n-1}3^{n+1}-2(3^n4^n)


    \displaystyle = 3^{n-1}4^{n-1}(4^2+3^2-2 \times 3 \times 4)

    How does that work?
  2. electriic_ink's Avatar
    2 05-07-2012  20:22
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    Re: (n+1)(n-1) Indices
    They factorised it using the laws of indices.
  3. anil10100's Avatar
    3 05-07-2012  20:36
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    Re: (n+1)(n-1) Indices
    (Original post by SubAtomic)
    Hi, this may be sixth form, not sure.

    How does this work, cannot quite fathom it, no help in book so am left in ambiguity.

    So, I am supposed to show that \displaystyle U_{n-1}U_{n+1}-U^2_n=12^{n-1} , for \displaystyle n=1,2,3...

    \displaystyle U_n=3^n+4^n

    \displaystyle U_{n-1}U_{n+1}=(3^{n-1}+4^{n-1})(3^{n+1}+4^{n+1}) \\ \\ =3^{2n}+3^{n-1}4^{n+1}+4^{n-1}3^{n+1}+4^{2n}

    fine


    \displaystyle U_n^2=(3^n+4^n)^2=3^{2n}+2(3^n4^ n)+4^{2n}

    fine


    Now, what I don't get is how this line becomes the next

    \displaystyle 3^{n-1}4^{n+1}+4^{n-1}3^{n+1}-2(3^n4^n)


    \displaystyle = 3^{n-1}4^{n-1}(4^2+3^2-2 \times 3 \times 4)

    How does that work?
    Just notice that 3^{n-1}4^{n+1}=3^{n-1}4^{n-1+2}=3^{n-1}4^{n-1}4^2 , similar thing for the others.
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  4. SubAtomic's Avatar
    4 05-07-2012  20:47
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    Re: (n+1)(n-1) Indices
    (Original post by anil10100)
    Just notice that 3^{n-1}4^{n+1}=3^{n-1}4^{n-1+2}=3^{n-1}4^{n-1}4^2 , similar thing for the others.
    Nice, will give it a whirl and see where I end up.
  5. SubAtomic's Avatar
    5 05-07-2012  21:43
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    Re: (n+1)(n-1) Indices
    So are these equivalent then

    \displaystyle a^{n-1}b^{n+1} \equiv (ab)^{n-1}b^2

    Seems to be according to the calculator.


    Back to the question, so

    \displaystyle 3^{n-1}4^{n-1+2}+4^{n-1}3^{n-1+2}-2(3^n4^n)


    \displaystyle = 3^{n-1}4^{n-1}4^2+3^{n-1}4^{n-1}3^2-2(3^n4^n)

    Is this right? What now? How do I get rid of those n in the bracket so it looks like the last bit of my OP?
  6. raheem94's Avatar
    6 05-07-2012  22:03
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    Re: (n+1)(n-1) Indices
    (Original post by SubAtomic)
    So are these equivalent then

    \displaystyle a^{n-1}b^{n+1} \equiv (ab)^{n-1}b^2

    Seems to be according to the calculator.
    In general, a^x \times a^y = a^{x+y}

    \displaystyle a^{n-1}b^{n+1} = a^{n-1} b^{ n - 1 + 2 } = a^{n-1} b^{ (n - 1) + 2 } = a^{n-1} b^{ (n - 1) } \times b^2 = (ab)^{n-1} b^2
  7. SubAtomic's Avatar
    7 05-07-2012  22:05
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    Re: (n+1)(n-1) Indices
    (Original post by raheem94)
    In general, a^x \times a^y = a^{x+y}

    \displaystyle a^{n-1}b^{n+1} = a^{n-1} b^{ n - 1 + 2 } = a^{n-1} b^{ (n - 1) + 2 } = a^{n-1} b^{ (n - 1) } \times b^2 = (ab)^{n-1} b^2
    Yeah that was my own little spin on things, not too sure on the second bit of that post.
    Last edited by SubAtomic; 05-07-2012 at 22:09.
  8. raheem94's Avatar
    8 05-07-2012  22:36
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    Re: (n+1)(n-1) Indices
    (Original post by SubAtomic)
    Yeah that was my own little spin on things, not too sure on the second bit of that post.
    Spoiler:
    Show

    3^{n-1} 4^{n+1} + 4^{n-1} 3^{n+1} - 2 (3^n 4^n ) = 3^{n-1} 4^{n+1} + 4^{n-1} 3^{n+1} - 2 (3 \times 4 )^n \\ = 3^{n-1} 4^{n+1} + 4^{n-1} 3^{n+1} - 2 (12)^n \\ = 3^n 3^{-1} 4^n 4^1 + 4^n 4^{-1} 3^n 3^1 - 2 \times 12^n = 3^n 4^n ( 4 \times 3^{-1} + 3 \times 4^{-1} ) - 2 \times 12^n \\ = (3 \times 4)^n \left( \frac43 + \frac34 \right) -2 \times 12^n = \frac{25}{12} 12^n - 2 \times 12^n \\ = 12^n \left( \frac{25}{12} -2 \right) = 12^n \times \frac1{12} = \frac{12^n}{12} = 12^{n-1}

    Last edited by raheem94; 05-07-2012 at 22:38.
  9. aznkid66's Avatar
    9 05-07-2012  23:04
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    Re: (n+1)(n-1) Indices
    (Original post by SubAtomic)
    Now, what I don't get is how this line becomes the next

    \displaystyle 3^{n-1}4^{n+1}+4^{n-1}3^{n+1}-2(3^n4^n)


    \displaystyle = 3^{n-1}4^{n-1}(4^2+3^2-2 \times 3 \times 4)

    How does that work?
    \displaystyle 3^{n-1}4^{n+1}+4^{n-1}3^{n+1}-2(3^n4^n)

    For easier comprehension, let's break it down by isolating the 3^{n-1}4^{n-1} in each individual term.



    First term:

    \displaystyle 3^{n-1}4^{n+1}

    Spoiler:
    Show
    Sub in n+1=n-1+2:

    3^{n-1}4^{(n-1)+2}

    Apply the rule a^(x+y)=(a^x)(a^y):

    3^{n-1}4^{n-1}4^2

    Factorize 3^(n-1)*4^(n-1) out (although it already is factorized because it's all multiplication here):


    (3^{n-1}4^{n-1})4^2



    Second term:

    4^{n-1}3^{n+1}

    Spoiler:
    Show
    Sub in n+1=(n-1)+2:

    4^{n-1}3^{(n-1)+2}

    Apply the rule a^(x+y)=(a^x)(a^y):

    4^{n-1}3^{n-1}3^2

    Factorize 3^(n-1)*4^(n-1) out (although it already is factorized because it's all multiplication here):


    (3^{n-1}4^{n-1})3^2



    Third term:

    -2(3^n4^n)

    Spoiler:
    Show
    Sub in n=(n-1)+1:

    -2(3^{(n-1)+1}4^{(n-1)+1})

    Apply the rule a^(x+y)=(a^x)(a^y)

    -2(3^{n-1}(3)4^{n-1}(4))

    Factorize 3^(n-1)*4^(n-1) out (although it already is factorized because it's all multiplication here):


    -(3^{n-1}4^{n-1})(2)(3)(4)



    All together now:

    (3^{n-1}4^{n-1})4^2+(3^{n-1}4^{n-1})3^2-(3^{n-1}4^{n-1})(2)(3)(4)

    Factorize out 3^(n-1)*4^(n-1):

    (3^{n-1}4^{n-1})(4^2+3^2-(2)(3)(4))
    Last edited by aznkid66; 05-07-2012 at 23:59. Reason: Silly typo.
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  10. SubAtomic's Avatar
    10 05-07-2012  23:54
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    Re: (n+1)(n-1) Indices
    (Original post by aznkid66)
    ...
    Nice explanation again, I thought of subbing the n=(n-1)+1 into the -2(3^n4^n) but thought it a bit of an odd way of doing things for some reason:rolleyes: you took it that extra step, thanks. You are missing the closing brackets in your third spoiler but I get ya


    (Original post by raheem94)
    ...
    Nice one mate, both your explanations have helped.


    Must take quite a while to be able to rep people again
    Last edited by SubAtomic; 05-07-2012 at 23:57.
  11. SubAtomic's Avatar
    11 06-07-2012  18:03
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    Re: (n+1)(n-1) Indices
    So let me see if I have got this right by trying another, tried one earlier and it went horribly wrong, I think.

    This first

    \displaystyle U_{n-4}U_{n+8}-U_n^2 for \displaystyle n=1,2,3...


    \displaystyle U_n=5^n+6^n


    \displaystyle U_{n-4}U_{n+8}=(5^{n-4}+6^{n-4})(5^{n+8}+6^{n+8}) \\ \\ =5^{2n+4}+5^{n-4}6^{n+8}+6^{n-4}5^{n+8}+6^{2n+4}

    Can I do something with 2^2 in the indices here? Seen as 4 and 8 are 2^2 and 2^3. Or would I be over-complicating things?

    Will carry on as per anyway,

    \displaystyle U_n^2=(5^n+6^n)^2=5^{2n}+2(5^n6^ n)+6^{2n}

    So


    \displaystyle U_{n-4}U_{n+8}-U_n^2=5^{2n+4}+5^{n-4}6^{n+8}+6^{n-4}5^{n+8}+6^{2n+4}-(5^{2n}+2(5^n6^n)+6^{2n})


    \displaystyle = 5^4+5^{n-4}6^{n-4}6^{12}+6^{n-4}5^{n-4}5^{12}+6^4-2(5^{n-4}6^{n-4})5^46^4


    \displaystyle = 5^{n-4}6^{n-4}(6^{12}+5^{12}-2 \times 5^4 \times 6^4)+5^4+6^4 \\ \\ = 30^{n-4}(6^{12}+5^{12}-2 \times 5^4 \times 6^4)+5^4+6^4


    \displaystyle=\dfrac{30^n(6^{12} +5^{12}-2 \times 5^4 \times 6^4)}{30^4}+5^4+6^4

    Is this right? If it is can I simplify further?
    Last edited by SubAtomic; 10-07-2012 at 12:32.
  12. raheem94's Avatar
    12 06-07-2012  18:40
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    Re: (n+1)(n-1) Indices
    (Original post by SubAtomic)
    So let me see if I have got this right by trying another, tried one earlier and it went horribly wrong, I think.

    This first

    \displaystyle U_{n-4}U_{n+8}-U_n^2 for \displaystyle n=1,2,3...


    \displaystyle U_n=5^n+6^n


    \displaystyle U_{n-4}U_{n+8}=(5^{n-4}+6^{n-4})(5^{n+8}+6^{n+8}) \\ \\ =5^{2n+4}+5^{n-4}6^{n+8}+6^{n-4}5^{n+8}+6^{2n+4}

    Can I do something with 2^2 in the indices here? Seen as 4 and 8 are 2^2 and 2^3. Or would I be over-complicating things?
    5^{ 2n + 4} = 5^{2n} \times 5^4 = (5^2)^n \times 5^4 = 25^n \times 625 = 625 (25)^n


    5^{n-4}6^{n+8}+6^{n-4}5^{n+8} = 5^n 5^{-4} 6^n 6^8 + 6^n 6^{-4} 5^n 5^8 = 5^n 6^n ( 5^{-4} 6^8 + 6^{-4} 5^8 ) \\ = ( 5 \times 6)^n ( 5^{-4} 6^8 + 6^{-4} 5^8 ) = 30^n (5^{-4} 6^8 + 6^{-4} 5^8 )


    6^{2n+4} = 6^{2n} \times 6^4 = (6^2)^n \times 6^4 = 1296(36)^n
    Last edited by raheem94; 06-07-2012 at 18:43.
  13. SubAtomic's Avatar
    13 06-07-2012  19:29
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    Re: (n+1)(n-1) Indices
    (Original post by raheem94)
    ...
    Yep I understand this, was meaning something like \displaystyle 5^{2n+4} \equiv 5^{2^1n+2^2} \equiv 5^{(n+2)2}

    \displaystyle 6^{n-4}5^{n+8} \equiv 6^{n-(2^2)}5^{n+(2^3)}

    Just trying stuff nothing major, seeing if a neater way.
    Last edited by SubAtomic; 06-07-2012 at 19:48.
  14. raheem94's Avatar
    14 06-07-2012  19:36
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    Re: (n+1)(n-1) Indices
    (Original post by SubAtomic)
    Yep I understand this, was meaning something like \displaystyle 5^{2n+4} \equiv 5^{2^1n+2^2} \equiv 5^{(n+2)2}

    \displaystyle 6^{n-4}5^{n+8} \equiv 6^{n-(2^2)}5^{n+(2^3)} \equiv 30^{(n-4)+(n+8)}

    Just trying stuff nothing major, seeing if a neater way.
    \displaystyle 5^{2n+4} \equiv 5^{2^1n+2^2} \equiv 5^{(n+2)2} \equiv (5^2)^{n+2} \equiv 25^{n+2} \equiv 25^n \times 25^2 = 625(25)^n
  15. raheem94's Avatar
    15 06-07-2012  19:43
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    Re: (n+1)(n-1) Indices
    (Original post by SubAtomic)
    \displaystyle 6^{n-4}5^{n+8} \equiv 6^{n-(2^2)}5^{n+(2^3)} \equiv 30^{(n-4)+(n+8)}
    This is wrong.

    6^{n-4}5^{n+8} \not\equiv 30^{(n-4)+(n+8)}

    If you want to check it, sub in a number.

    \displaystyle \text{Let } n = 1, \\ 6^{n-4}5^{n+8} = 6^{-3}5^9 = 9042.24537 \\ 30^{(n-4)+(n+8)} = 30^{-3 + 9 } = 30^6 =729000000
  16. SubAtomic's Avatar
    16 06-07-2012  19:48
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    Re: (n+1)(n-1) Indices
    (Original post by raheem94)
    ..
    Edited, experiments and all.
  17. SubAtomic's Avatar
    17 06-07-2012  19:52
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    Re: (n+1)(n-1) Indices
    (Original post by raheem94)
    ...
    The main thing I was wondering about is how the indices have common factors and if they can be simplified in a certain fashion, not actually quite sure what I was thinking but it was something lol.
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