题目内容
等比数列{an}的首项为a,公比为q,Sn为其前n项和,求S=S1+S2+…+Sn.
解:当q=1时,an=q,Sn=na1,∴S=a+2a+…+na=(1+2+…+n)a=
a;
当q≠1时,Sn=
.
∴S=S1+S2+…+Sn
=
[(1-q)+(1-q2)+…+(1-qn)]
=
[n-(q+q2+…+qn)]
=
[n-
]
=
-
.
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题目内容
等比数列{an}的首项为a,公比为q,Sn为其前n项和,求S=S1+S2+…+Sn.
解:当q=1时,an=q,Sn=na1,∴S=a+2a+…+na=(1+2+…+n)a=
a;
当q≠1时,Sn=
.
∴S=S1+S2+…+Sn
=
[(1-q)+(1-q2)+…+(1-qn)]
=
[n-(q+q2+…+qn)]
=
[n-
]
=
-
.