求 1/2+1/(2*3)+1/(3*4)+1/(4*5)+……+1/(2008*2009)的值

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求 1/2+1/(2*3)+1/(3*4)+1/(4*5)+……+1/(2008*2009)的值
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求 1/2+1/(2*3)+1/(3*4)+1/(4*5)+……+1/(2008*2009)的值
求 1/2+1/(2*3)+1/(3*4)+1/(4*5)+……+1/(2008*2009)的值

求 1/2+1/(2*3)+1/(3*4)+1/(4*5)+……+1/(2008*2009)的值
1/2+1/(2*3)+1/(3*4)+1/(4*5)+……+1/(2008*2009)
=1-1/2+1/2-1/3+1/3-1/4+1/5-……+1/2008-1/2009
=1-1/2009
=2008/2009
希望杯的题来吧?

用裂项公式做。1/(n*(n+1)=1/n-1/(n+1).代入后经过抵消就只剩个1/2009啦