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Math Help - [SOLVED] Proof

  1. #1
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    [SOLVED] Proof

    I am kinda new to proof and can find it difficult sometimes. Can some help me to solves the following proofs.

    1. d/dx (x^n)= nx^n-1


    2. r(r+1) = 1/3 n(n+1)(n+2)


    3. 1.1!+ 2.2! + 3.3! +...+ n.n! = (n+1)!-1


    Thanks
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  2. #2
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    Quote Originally Posted by maths2008 View Post
    I am kinda new to proof and can find it difficult sometimes. Can some help me to solves the following proofs.
    These all look like "proof by induction" problems. Prove the statement is true for n= 1 (the first does not say that n is a positive integer but I am assuming that), the prove "if the statement is true for n=k then it is true for n= k+1.

    1. d/dx (x^n)= nx^n-1
    If n= 1, this says d(x^1)/dx= 1(x^0). That is true because y= x is a linear function with slope 1= 1(x^0).

    Suppose you know that d/dx (x^k)= k x^k-1. x^k+1= x(x^k). Differentiate that using the product rule.

    2. r(r+1) = 1/3 n(n+1)(n+2)
    I don't know what you mean here. Assuming you really mean [tex]r^k(r)(r+1)[/itex] this is NOT true even for n= 1. If you mean \sum_{k=1}^n k(k+1), when n= 1 the left sides is just 1(2)= 2 and the right sides is (1/3)(1)(2)(3)= 2 so it is true for n= 1.
    Suppose it is true for n= i:
    \sum_{k=1}^i k(k+1)= (1/3)i(i+1)(i+2)
    Then [tex]\sum_{k=1}^{i+1} n(n+1)= \sum_{k=1}^i k(k+1) + (i+1)(i+2)[/itex]


    3. 1.1!+ 2.2! + 3.3! +...+ n.n! = (n+1)!-1
    When n= 1, the left sides is 1.1!= 1 and the right side is 2!- 1= 1 so it is true for n= 1.
    Assume 1.1!+ 2.2! + 3.3! +...+ k.k! = (k+1)!-1. Then 1.1!+ 2.2! + 3.3! +...+ (k+1).(k+1)! = (k+1)!-1 + (k+1).(k+1)! Try factoring out (k+1)!

    Thanks[/QUOTE]
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  3. #3
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    thanks for the help
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