已知数列{an}是公差不为零的等差数列且a1=1,且a2,a4,a8成等比数列.①求通项an
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已知数列{an}是公差不为零的等差数列且a1=1,且a2,a4,a8成等比数列.①求通项an
②若bn=1/[an*a(n+1)],求{bn}的前n项和Sn
②若bn=1/[an*a(n+1)],求{bn}的前n项和Sn
![已知数列{an}是公差不为零的等差数列且a1=1,且a2,a4,a8成等比数列.①求通项an](/uploads/image/z/15301103-23-3.jpg?t=%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97%EF%BD%9Ban%7D%E6%98%AF%E5%85%AC%E5%B7%AE%E4%B8%8D%E4%B8%BA%E9%9B%B6%E7%9A%84%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%E4%B8%94a1%3D1%2C%E4%B8%94a2%2Ca4%2Ca8%E6%88%90%E7%AD%89%E6%AF%94%E6%95%B0%E5%88%97.%E2%91%A0%E6%B1%82%E9%80%9A%E9%A1%B9an)
设an=a1+(n-1)d
即an=1+(n-1)d
a2,a4,a8成等比数列
(1+3d)^2=(1+d)(1+7d)
1+6d+9d^2=1+8d+7d^2
2d^2=2d
d≠0 d=1
an=1+(n-1)=n
bn=1/[n(n+1)]
Sn=1/(1*2)+1/(2*3)+1/(3*4)+.+ 1/[n(n+1)]
=1/1-1/2+1/2-1/3+1/3-1/4+.+1/n-1/(n+1)
=1-1/(n+1)=n/(n+1)
即an=1+(n-1)d
a2,a4,a8成等比数列
(1+3d)^2=(1+d)(1+7d)
1+6d+9d^2=1+8d+7d^2
2d^2=2d
d≠0 d=1
an=1+(n-1)=n
bn=1/[n(n+1)]
Sn=1/(1*2)+1/(2*3)+1/(3*4)+.+ 1/[n(n+1)]
=1/1-1/2+1/2-1/3+1/3-1/4+.+1/n-1/(n+1)
=1-1/(n+1)=n/(n+1)
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