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Math Help - Prove that the sum of fourth powers of the first n integers is 1/30n(n+1) (2n+1)(

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    Prove that the sum of fourth powers of the first n integers is 1/30n(n+1) (2n+1)(

    Prove that the sum of fourth powers of the first n integers is 1/30n(n+1) (2n+1)(3n2+3n+1)
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by rajeev000 View Post
    Prove that the sum of fourth powers of the first n integers is 1/30n(n+1) (2n+1)(3n2+3n+1)
    sort of confused about the formula you wrote, since i am sure there are brackets missing that would be needed to clarify. nonetheless, induction would be the way to go here.
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    Sum of fourth powers: proof by induction

    Hello rajeev000 -

    Quote Originally Posted by rajeev000 View Post
    Prove that the sum of fourth powers of the first n integers is 1/30n(n+1) (2n+1)(3n2+3n+1)
    The formula you're trying to prove is:

    \sum_{i=1}^n n^4=\frac{1}{30}n(n+1)(2n+1)(3n^2+3n-1)

    Prove by induction. Assume that this is true for some n. Then:

    \sum_{i=1}^{n+1} n^4=\frac{1}{30}n(n+1)(2n+1)(3n^2+3n-1) + (n+1)^4

    When you're faced with an expression like this, don't multiply out all the brackets. Look for common factors. Here we have a common factor (n+1). So factorise:

    \sum_{i=1}^{n+1} n^4=\frac{1}{30}(n+1)(n(2n+1)(3n^2+3n-1)+30(n+1)^3)

    Then you can multiply out within the long bracket, and simplify. You will then get
    (I'll leave you to check this out):

    \sum_{i=1}^{n+1} n^4=\frac{1}{30}(n+1)(64n^4+39n^3+91n^2+89n+30)

    Don't be afraid of it! You know that we're looking for an expression which is the equivalent of the original one, with
    n replaced by (n+1). So we should have a factor now of (n+2) and another one of (2n+3). So take out these factors, and you get (again, I'll leave you to supply the missing lines):

    \sum_{i=1}^{n+1} n^4=\frac{1}{30}(n+1)(n+2)(2n+3)(3n^2+9n+5)

    All that remains is to show that the final bracket is the same as (3(n+1)^2+3(n+1)-1) - which it is - and to show that the formula is true when n=1. Can you do this now?

    Grandad

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