sum of the reciprocals of the squares of the consecutive even integers , n-2 ,n and n+2?

Find the rational expression, in its simplest form, the represents the sum of the reciprocals of the squares of the consecutive even integers , n-2 ,n and n+2

5 Answers

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  • Anonymous
    1 decade ago
    Best answer

    The sum of the reciprocals of the squares is:

    1 / (n - 2)² + 1 / n² + 1 / (n + 2)²

    Get common denominators and add:

    [n²(n + 2)² + (n - 2)²(n + 2)² + n²(n - 2)²] / [n²(n - 2)²(n + 2)²]

    [n²(n² + 4n + 4) + (n² - 4n + 4)(n² + 4n + 4) + n²(n² - 4n + 4)] / [n²(n - 2)²(n + 2)²]

    [(n^4 + 4n³ + 4n²) + (n^4 + - 8n² + 16) + (n^4 - 4n³ + 4n²)] / [n²(n - 2)²(n + 2)²]

    (3n^4 + 16) / [n²(n - 2)²(n + 2)²]

  • Anonymous
    1 decade ago

    Squares of the consecutive even integers , n-2 ,n and n+2:

    (n - 2)^2

    n^2

    (n + 2)^2

    Reciprocals:

    1 / (n - 2)^2

    1 / (n)^2

    1 / (n + 2)^2

    For sum, find a common denominator:

    [(n)(n + 2)]^2 / [(n - 2)(n)(n + 2)] ^ 2

    +

    [(n + 2)(n - 2)]^2 / [(n - 2)(n)(n + 2)] ^ 2

    +

    [(n)(n - 2)]^2 / [(n - 2)(n)(n + 2)] ^ 2

    =

    [(n^2 + 2n)]^2 / [(n - 2)(n)(n + 2)] ^ 2

    +

    [(n^2 -4)]^2 / [(n - 2)(n)(n + 2)] ^ 2

    +

    [(n^2 - 2n)]^2 / [(n - 2)(n)(n + 2)] ^ 2

    =

    (n^4 + 4n^3 + 4n^2) / [(n - 2)(n)(n + 2)] ^ 2

    +

    (n^4 -8n^2 + 16) / [(n - 2)(n)(n + 2)] ^ 2

    +

    (n^4 -4n^3 + 4n^2)] / [(n - 2)(n)(n + 2)] ^ 2

    =(3n^4 + 16) / [(n^3 - 4n)] ^ 2

    =(3n^4 + 16) / (n^6 - 8n^4 +16n^2)

    =(3n^4 + 16) / (n^2)(n^4 - 8n^2 +16)

    =(3n^4 + 16) / [(n)(n^2 - 4)]^2

  • 3 years ago

    let the consecutive integers be x , x+a million a million/x + a million/ x+a million = 7/24 (x +a million ) + x / x(x+a million ) = 24( x + x +a million ) = 7 (x^2 + x ) 24(2x +a million) = 7x^2 +7x 48x +24 = 7x^2 + 7x 7x^2 - 40-one x - 24 =0 x can't be an integer. thankyou 7x^2 - 40-one x - 24

  • 1 decade ago

    The answer is

    1/(n^2-4n+4)+1/(n^2)+1/(n^2+4n+4)

    Not simplifying beyond that...

    Hope this helps...

    Source(s): Me
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  • 1 decade ago

    (3n^4+16) / [ (n^2)*(n-2)^2*(n+2)^2]

    Source(s): algebra
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