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](/uploads/image/z/8849636-44-6.jpg?t=%E6%B1%82+1%2F2%2B1%2F%282%2A3%29%2B1%2F%283%2A4%29%2B1%2F%284%2A5%29%2B%E2%80%A6%E2%80%A6%2B1%2F%282008%2A2009%29%E7%9A%84%E5%80%BC)
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