已知数列{An}的通项公式An=n.cos(nπ/2+π/3),记Sn=a1+a2+…+an,求S2002

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a(n) = ncos(nπ/2+π/3),n = 1,2,...
a(4k) = 4kcos(4kπ/2+π/3) = 4kcos(π/3) = 2k,
a(4k+1) = (4k+1)cos[(4k+1)π/2+π/3] = (4k+1)cos[π/2+π/3]
= (4k+1)sin(-π/3) = -(4k+1)3^(1/2)/2.
a(4k+2) = (4k+2)cos[(4k+2)π/2+π/3] = (4k+2)cos[π+π/3]
= -(4k+2)/2 = -(2k+1)
a(4k+3) = (4k+3)cos[(4k+3)π/2+π/3] = (4k+3)cos[3π/2+π/3]
= (4k+3)sin(-π/3) = -(4k+3)3^(1/2)/2.
S(2002) = [a(1) + a(5) + ...+ a(2001)] + [a(2) + a(6) + ...+ a(2002)] + [a(3) + a(7) + ...+ a(1999)] + [a(4) + a(8) + ...+ a(2000)]
= -3^(1/2)/2[1 + 5 + ...+ 2001] - [1 + 3 + ...+ 1001] -3^(1/2)/2[3 + 7 + ...+ 1999] + [2 + 4 + ...+ 1000]
= -3^(1/2)/2[1 + 3 + 5 + ...+ 1999 + 2001] - [1 + 1001]^2/4 + [2 + 1000]*500/2
= -3^(1/2)/2[1 + 2001]^2/4 - 1002^2/4 + 501*500
= -3^(1/2)/2[2002]^2/4 - 501^2 + 501*500
= -3^(1/2)/2[1001]^2 - 501

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