# 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

- Anonymous1 decade agoFavourite 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]

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- Anonymous1 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

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- 4 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

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- 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...

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- HeavySnarkerLv 51 decade ago
(3n^4+16) / [ (n^2)*(n-2)^2*(n+2)^2]

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