已知数列an=(n+1)×(910)n,求{an}的前n项和Sn.-数学

题目简介

已知数列an=(n+1)×(910)n,求{an}的前n项和Sn.-数学

题目详情

已知数列an=(n+1)×(
9
10
)n,求{an}的前n项和Sn
题型:解答题难度:中档来源:不详

答案

∵an=n+1为等差数列,bn=(class="stub"9
10
)
n
为等比数列
Sn=2×class="stub"9
10
+3×(class="stub"9
10
)
2
+…+(n+1)•(class="stub"9
10
)
n

class="stub"9
10
Sn
=2×(class="stub"9
10
)
2
+3×(class="stub"9
10
)
3
+…+(n+1)×(class="stub"9
10
)
n

两式相减,class="stub"1
10
Sn
=class="stub"9
5
+[(class="stub"9
10
)
2
+(class="stub"9
10
)
3
+…+(class="stub"9
10
)
n
]
-(n+1)•(class="stub"9
10
)
n+1

=class="stub"9
5
+class="stub"81
10
×[1- (class="stub"9
10
)
n
 ]
-(n+1)×(class="stub"9
10
)
n+1

=class="stub"99
10
-(class="stub"9
10
)
n+1
(n+10)

Sn=99-9(n+10)×(class="stub"9
10
)
n

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