n(n+1)/1+(n+1)(n+2)/1+.+(n+2007)(n+2008)/1=?
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n(n+1)/1+(n+1)(n+2)/1+.+(n+2007)(n+2008)/1=?n(n+1)/1+(n+1)(n+2)/1+.+(n+2007)(n+2008)/1=?n(n+1)/1+(n+
n(n+1)/1+(n+1)(n+2)/1+.+(n+2007)(n+2008)/1=?
n(n+1)/1+(n+1)(n+2)/1+.+(n+2007)(n+2008)/1=?
n(n+1)/1+(n+1)(n+2)/1+.+(n+2007)(n+2008)/1=?
用裂项法：
1/(n+1)(n+2)=1/(n+1)-1/(n+2)
各项同相处理,再相加,抵消得：
所以其和=1/(n+1)-1/(n+2008)
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