1.寫出它的通項(xiàng)an,并證明數(shù)列{an}是等差數(shù)列
an=1+1/n+2/n+.+(n-1)/n=1/n+2/n+.+n/n=(1+n)/2;
a(n+1)-an
=((n+2)/2-(n+1)/2=1/2
2.設(shè) bn=1/an*an-1,求數(shù)列{bn}的前n項(xiàng)和
這個(gè)式子書寫不規(guī)范,我只能猜.
b1=1/（a1*a0）,不存在.
b2=1/(a2*a1)=4/[(2+1)*2]=4*(1/2-1/3)
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bn=1/(an*an-1)=4/[(n+1)*n]=4*[1/n-1/(n+1)]
數(shù)列{bn}的前n項(xiàng)和
=4*(1/2-1/3)+4*(1/3-1/4)+.+4*[1/n-1/(n+1)]
=4*[1/2-1/(n+1) ]