If n satisfies the square of (n-2006) + the square of (2008-n) = 1, then (n-2006) (2008-n) is equal to?

If n satisfies the square of (n-2006) + the square of (2008-n) = 1, then (n-2006) (2008-n) is equal to?


Square of (n-2006) + square of (2008-n)
=Square of (n-2006) + square of (2008-n) + 2 (n-2006) (2008-n) - 2 (n-2006) (2008-n)
=(n-2006+2008-n)²-2（n-2006)(2008-n)
=4-2（n-2006)(2008-n)
=1
（n-2006）（2008-n）=（4-1）/2=3/2



（1/1+2）+（1/1+2+3）+…… +（1/1+2+3+4+…… +What is the equivalent of 2008?


The general term formula is 2 / (n + 1) (n + 2), so the sum is 2 (1 / 2-1 / N + 2), that is 1-2 / N + 2

Elden Ring Premiere

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