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Indices Help [Factorisation & Simplication] watch

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    (Original post by DylanJ42)
    so we have  \displaystyle 2^n (2n)

    If you had  \displaystyle a =bcd you could write this as  \displaystyle a =cbd = dbc = bdc etc etc, ie it doesnt matter what order the letters are in.

    Using this, we can rewrite the first expression as  \displaystyle (2)2^n(n)

     \displaystyle 2 \times 2^n = 2^{n+1}

     \displaystyle \therefore  2 \times 2^n \times n = 2^{n+1} \times n
    I understand this completely now, thank you very much for your help!

    (Original post by ZXLPK42)
    I understand this completely now, thank you very much for your help!
    my pleasure
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