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(n+1)(n-1) Indices

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    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?
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    They factorised it using the laws of indices.
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    (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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    (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.
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    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?
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    (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
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    (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.
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    (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}

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    (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))
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    (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
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    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?
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    (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
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    (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.
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    (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
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    (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
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    (Original post by raheem94)
    ..
    Edited, experiments and all.
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    (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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