计算(x+1)(x+2)/1+(x+2)(x+3)/1+(x+3)(x+4)/1+……+(x+2008)(x+2009)
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计算(x+1)(x+2)/1+(x+2)(x+3)/1+(x+3)(x+4)/1+……+(x+2008)(x+2009)/1
RT
RT
![计算(x+1)(x+2)/1+(x+2)(x+3)/1+(x+3)(x+4)/1+……+(x+2008)(x+2009)](/uploads/image/z/16984257-33-7.jpg?t=%E8%AE%A1%E7%AE%97%28x%2B1%29%28x%2B2%29%2F1%2B%28x%2B2%29%28x%2B3%29%2F1%2B%28x%2B3%29%28x%2B4%29%2F1%2B%E2%80%A6%E2%80%A6%2B%28x%2B2008%29%28x%2B2009%29)
拆项法,应用公式1/[n(n+1)]=1/n-1/(n+1):
原式=1/(x+1)-1/(x+2)+1/(x+2)-1/(x+3)+...+1/(x+2008)-1/(x+2009)
=1/(x+1)-1/(x+2009)
=2008/[(x+1)(x+2009)]
原式=1/(x+1)-1/(x+2)+1/(x+2)-1/(x+3)+...+1/(x+2008)-1/(x+2009)
=1/(x+1)-1/(x+2009)
=2008/[(x+1)(x+2009)]
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