已知数列{an}是公差不为0的等差数列,a1=2,且a2,a3,a4+1成等比数列.(Ⅰ)求数列{an}的通项公式;(Ⅱ)设bn=2n•(an+2),求数列{bn}的前n项和Sn.-高二数学

题目简介

已知数列{an}是公差不为0的等差数列,a1=2,且a2,a3,a4+1成等比数列.(Ⅰ)求数列{an}的通项公式;(Ⅱ)设bn=2n•(an+2),求数列{bn}的前n项和Sn.-高二数学

题目详情

已知数列{an}是公差不为0的等差数列,a1=2,且a2,a3,a4+1成等比数列.
(Ⅰ)求数列{an}的通项公式;
(Ⅱ)设bn=
2
n•(an+2)
,求数列{bn}的前n项和Sn
题型:解答题难度:中档来源:不详

答案

(Ⅰ)设数列{an}的公差为d,由a1=2和a2,a3,a4+1成等比数列,得
(2+2d)2-(2+d)(3+3d),解得d=2,或d=-1,
当d=-1时,a3=0,与a2,a3,a4+1成等比数列矛盾,舍去.
∴d=2,
∴an=a1+(n-1)d=2+2(n-1)=2n.
即数列{an}的通项公式an=2n;
(Ⅱ)由an=2n,得
bn=class="stub"2
n•(an+2)
=class="stub"2
n(2n+2)
=class="stub"1
n(n+1)
=class="stub"1
n
-class="stub"1
n+1

∴Sn=b1+b2+b3+…+bn
=1-class="stub"1
2
+class="stub"1
2
-class="stub"1
3
+class="stub"1
3
-class="stub"1
4
+…+class="stub"1
n
-class="stub"1
n+1
=class="stub"n
n+1

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