题目内容
以数列{an}的任意两项为坐标的点Pn(an,an+1)(n∈N*)均在一次函数y=2x+8的图象上,数列{bn}满足条件:bn=an+1-an(n∈N*,b1≠0)且a1=1.(文)求数列{bn}的前n项和Tn.
(理)求数列{an}的前n项和Sn和数列{bn}的前n项和Tn.
解析：(文)由题意an+1=2an+8,
bn=an+1-an=an+8.
bn+1=an+1+8=2an+16.∴
=2.
b1=a2-a1=2a1+8-a1=a1+8=9.
∴bn=9×2n-1.
Tn=
=
=9×(2n-1).
(理)由题意an+1=2an+8,
bn=an+1-an=an+8.
bn+1=an+1+8=2an+16.∴
=2.
b1=a2-a1=2a1+8-a1=a1+8=9.
∴bn=9×2n-1.
Tn=
=
=9×(2n-1).
由bn=an+8,得an=bn-8=9×2n-1-8,
Sn=(9-8)+(9×2-8)+(9×22-8)+…+(9×2n-1-8)
=9(1+2+22+…+2n-1)-8n
=
-8n
=9×(2n-1)-8n
=9×2n-8n-9.
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