[No correct answers were submitted for this problem.]
We will use the method of partial fractions. We take the fraction 1⁄(n2-4) and note that the denominator factors as (n+2)(n-2). We will thus try to rewrite the fraction as the sum of a fraction with denominator n+2 and a fraction with denominator n-2.
So we want to find numerators A and B such that:
1⁄(n2-4) = A⁄(n+2) + B⁄(n-2)
Multiplying both sides by (n+2)(n-2), we obtain:
1 = A(n-2) + B(n+2)
Rearranging terms, we have:
1 = (A+B)n + 2(B-A)
This identity needs to hold for all values of n. So we need to have:
A + B = 0
2(B - A) = 1
Solving this pair of equations, we get A = -1/4 and B = 1/4.
So 1⁄(n2-4) = (1⁄(n-2) - 1⁄(n+2))/4.
Now, the sum we are trying to approximate is:
1000(1⁄(32-4) + 1⁄(42-4) + 1⁄(52-4) + ... + 1⁄(100002-4))
Using our partial fraction decomposition, we can rewrite this as:
1000((1⁄(3-2) - 1⁄(3+2))/4 + (1⁄(4-2) - 1⁄(4+2))/4 + (1⁄(5-2) - 1⁄(5+2))/4 + ... + (1⁄(10000-2) - 1⁄(10000+2))/4)
Factoring out the common 1/4 factor, we obtain:
250(1⁄(3-2) - 1⁄(3+2) + 1⁄(4-2) - 1⁄(4+2) + 1⁄(5-2) - 1⁄(5+2) + ... + 1⁄(10000-2) - 1⁄(10000+2))
= 250(1 - 1⁄5 + 1⁄2 - 1⁄6 + 1⁄3 - 1⁄7 + ... + 1⁄9998 - 1⁄10002)
The series inside the parentheses is an example of a telescoping series. The same terms show up both positive and negative, and cancel out. Examining the series, we see that the positive terms 1, 1/2, 1/3, and 1/4 do not cancel out, and the negative terms -1/9999, -1/10000, -1/10001, and -1/10002 do not cancel out. All other terms do. So our expression reduces to:
250(1 + 1/2 + 1/3 + 1/4 - 1/9999 - 1/10000 - 1/10001 - 1/10002)
1 + 1/2 + 1/3 + 1/4 = 2.083. Multiplying this by 250 gives a bit more than 520.8. From this, we have to subtract 250 times (1/9999 + 1/10000 + 1/10001 + 1/10002). But this is approximately 1000(1/10000) = .1, so subtracting it will leave roughly 520.7. We want the nearest integer, which will be 521.
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