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Math Help - show by differentiation

  1. #1
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    show by differentiation

    if (1 + x)^n = (n/0) + (n/1)x + (n/2)x^2 + ........... + (n/n)x^n
    show by differentiation that

    2(2/n) + (2x3)(n/3) + (3x4)(n/4) + ......... + n(n-1)(n/n) = n(n-1)2^n-2
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  2. #2
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    Quote Originally Posted by deej813 View Post
    if (1 + x)^n = (n/0) + (n/1)x + (n/2)x^2 + ........... + (n/n)x^n
    show by differentiation that

    2(2/n) + (2x3)(n/3) + (3x4)(n/4) + ......... + n(n-1)(n/n) = n(n-1)2^n-2
    You mean (1+ x)^n= \left(\begin{array}{c}n \\ 0\end{array}\right)+ \left(\begin{array}{c}n \\ 1\end{array}\right)x+ \left(\begin{array}{c}n \\ 2\end{array}\right)x^2+ \cdot\cdot\cdot +\left(\begin{array}{c}n \\ n\end{array}\right)x^n
    (what you wrote looks too much like fractions!)

    Have you tried this at all? The " 2^{n-2}" on the right makes me think of differentiating twice (so that the power is reduced to n-2) and then taking x= 1 (so that x+1= 1+ 1= 2). What do you get if you do that on both sides?
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  3. #3
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    yes thats what i mean
    thanks
    i don kow how to write them like that
    um no i still dont get it
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  4. #4
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    Quote Originally Posted by deej813 View Post
    if (1 + x)^n = (n/0) + (n/1)x + (n/2)x^2 + ........... + (n/n)x^n
    show by differentiation that

    2(2/n) + (2x3)(n/3) + (3x4)(n/4) + ......... + n(n-1)(n/n) = n(n-1)2^n-2
    The equation we have is:

    (1+x)^n = \sum_{k=0}^n \binom{n}{k}x^k 1^{n-k} = \binom{n}{0}x^0 + \binom{n}{1}x + ... +\binom{n}{n-1}x^{n-1}+ \binom{n}{n}x^n

    Differentiating both sides once, we get:

    n(x+1)^{n-1} = \binom{n}{1} + 2\binom{n}{2}x + ... + (n-1)\binom{n}{n-1}x^{n-2} + n\binom{n}{n}x^{n-1}

    Differentiating again, we get:

    n(n-1)(x+1)^{n-2}= 2\binom{n}{2} + 2 \cdot 3\binom{n}{3}x + ... + (n-1)(n-2)\binom{n}{n-1}x^{n-3} + n(n-1)\binom{n}{n}x^{n-2}

    If we take x = 1, we get:

    2\binom{n}{2} + 2 \cdot 3\binom{n}{3} + ... + (n-1)(n-2)\binom{n}{n-1} + n(n-1)\binom{n}{n} = n(n-1)(1+1)^{n-2} = n(n-1)2^{n-2}
    Last edited by Defunkt; September 1st 2009 at 01:50 PM.
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  5. #5
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    Quote Originally Posted by deej813 View Post
    yes thats what i mean
    thanks
    i don kow how to write them like that
    um no i still dont get it
    Do you not know that the derivative of x^n is n x^{n-1}? (For n not equal to -1.) That's usually the first derivative you learn!
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  6. #6
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    yeah i do get that
    um i get most of it
    just after the first differentiation where does the +(n-1)(n/n-1)x^n-2 come from
    sorry i wrote it like fractions again
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  7. #7
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    Quote Originally Posted by deej813 View Post
    yeah i do get that
    um i get most of it
    just after the first differentiation where does the +(n-1)(n/n-1)x^n-2 come from
    sorry i wrote it like fractions again
    What's the derivative of \binom{n}{k}x^t, with respect to x?
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  8. #8
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    isnt it
    t(n/k)x^t-1
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  9. #9
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    Quote Originally Posted by deej813 View Post
    isnt it
    t(n/k)x^t-1
    Correct. The term you asked about - I just did not specifically write it in the initial equation. Of course, this does not mean it was not there!

    I fixed it now so it is clearer. Post here if you still need clarification!
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  10. #10
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    ok great
    i get it now
    thanks heaps
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