已知:数列{an}的前n项和Sn=n2+2n(n∈N*) (1)求:通项 an (2)求和:1/a1a2+1/a2a3+1/a3a4+…+1/anan+1.
已知:数列{an}的前n项和Sn=n2+2n(n∈N*)
(1)求:通项 an
(2)求和:
+
+
+…+
.
(1)求:通项 an
(2)求和:
1 |
a1a2 |
1 |
a2a3 |
1 |
a3a4 |
1 |
anan+1 |
数学人气:398 ℃时间:2020-02-03 18:28:14
优质解答
(1)∵a1=S1=3,∴当n≥2时,an=Sn-Sn-1=2n+1,当n=1时,a1=3,∴an=2n+1…(6分)(2)当n=1时,原式=130当n≥2时,1anan+1=1(2n+1)(2n+3)=12•(12n+1−12n+3)∴原式=130+12•(15−17+…+12n+1−12n+3)=130+12(1...
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