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简便算法数学题
1/(2^2)+2/(2^3)+3(2^4)+……+2001/(2^2002)
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人气:144 ℃ 时间:2020-04-01 13:55:07
解答
设S=1/(2^2)+2/(2^3)+3/(2^4)+……+2000/(2^2001)+2001/(2^2002) ①
则2S=1/2+2/(2^2)+3/(2^3)+4/(2^4)+……+2001/(2^2001)+2001/(2^2002) ②
①-②得
-S=1/2-1/(2^2)-1/(2^3)-4/(2^4)-……-1/(2^2001)-2001/(2^2002)
==1-(1/2+1/(2^2)+1/(2^3)+……+1/(2^2001))-2001/(2^2002)
==1-(1/2)*(1-(1/2)^2000)/(1-1/2)-2001/(2^2002)
==1/(2^2000)-2001/(2^2002)
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