等差数列{an}中,a2=8,S6=66(1)求数列{an}的通项公式an;(2)设bn=2(n+1)an,Tn=b1+b2+b3+…+bn,求Tn.-数学

题目简介

等差数列{an}中,a2=8,S6=66(1)求数列{an}的通项公式an;(2)设bn=2(n+1)an,Tn=b1+b2+b3+…+bn,求Tn.-数学

题目详情

等差数列{an}中,a2=8,S6=66
(1)求数列{an}的通项公式an
(2)设bn=
2
(n+1)an
,Tn=b1+b2+b3+…+bn,求Tn
题型:解答题难度:中档来源:不详

答案

(1)设等差数列{an}的公差为d,则有
a1+d=8
6a1+15d=66
        …(2分)
解得:a1=6,d=2,…(4分)
∴an=a1+d(n-1)=6+2(n-1)=2n+4             …(6分)
(2)bn=class="stub"2
(n+1)an
=class="stub"1
(n+1)(n+2)
=class="stub"1
n+1
-class="stub"1
n+2
        …(9分)
∴Tn=b1+b2+b3+…+bn=class="stub"1
2
-class="stub"1
3
+class="stub"1
3
-class="stub"1
4
+…+class="stub"1
n+1
-class="stub"1
n+2
=class="stub"1
2
-class="stub"1
n+2
=class="stub"n
2n+4
                                   …(12分)

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