1/n(n+k)
=1/k*k/n(n+k)
=1/k*[(n+k)-n]/n(n+k)
=1/k*[(n+k)/n(n+k)-n/n(n+k)]
=1/k[1/n-1/(n+k)]
所以1/1*4+1/4*7+1/7*10+······+1/2008*2011
=1/3*(1-1/4+1/4-1/7+1/7-1/10+……+1/2008-1/2011)
=1/3*(1-1/2011)
=670/2011
1/n(n+k)
=1/k*k/n(n+k)
=1/k*[(n+k)-n]/n(n+k)
=1/k*[(n+k)/n(n+k)-n/n(n+k)]
=1/k[1/n-1/(n+k)]
所以1/1*4+1/4*7+1/7*10+······+1/2008*2011
=1/3*(1-1/4+1/4-1/7+1/7-1/10+……+1/2008-1/2011)
=1/3*(1-1/2011)
=670/2011