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求和:1+
4
5
+
7
52
+…+
3n−2
5n−1
人气:381 ℃ 时间:2020-03-28 07:07:48
解答
设Sn=1+
4
5
+
7
52
+…+
3n−5
5n−2
+
3n−2
5n−1
   ①
1
5
Sn=
1
5
+
4
52
+
7
53
+…+
3n−5
5n−1
+
3n−2
5n
     ②
①-②得:
4
5
Sn=1+
3
5
+
3
52
+…+
3
5n−1
3n−2
5n

=1+3×
1
5
(1−
1
5n−1
)
1−
1
5
3n−2
5n

=
5n−12n−7
5n

∴Sn=
5n−12n−7
16×5n−1
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