(1)已知数列{an}的前n项和Sn=3n2-2n,求证数列{an}成等差数列.(2)已知1a,1b,1c成等差数列,求证b+ca,c+ab,a+bc也成等差数列.-数学

题目简介

(1)已知数列{an}的前n项和Sn=3n2-2n,求证数列{an}成等差数列.(2)已知1a,1b,1c成等差数列,求证b+ca,c+ab,a+bc也成等差数列.-数学

题目详情

(1)已知数列{an}的前n项和Sn=3n2-2n,求证数列{an}成等差数列.
(2)已知
1
a
1
b
1
c
成等差数列,求证
b+c
a
c+a
b
a+b
c
也成等差数列.
题型:解答题难度:中档来源:不详

答案

(1)证明:当n=1时,a1=S1=3-2=1,
当n≥2时,an=Sn-Sn-1=3n2-2n-[3(n-1)2-2(n-1)]=6n-5,
n=1时,亦满足,∴an=6n-5(n∈N*).
首项a1=1,an-an-1=6n-5-[6(n-1)-5]=6(常数)(n∈N*),
∴数列{an}成等差数列且a1=1,公差为6.
(2)∵class="stub"1
a
class="stub"1
b
class="stub"1
c
成等差数列,
class="stub"2
b
=class="stub"1
a
+class="stub"1
c
化简得2ac=b(a+c).
class="stub"b+c
a
+class="stub"a+b
c
=
bc+c2+a2+ab
ac
=
2ac+a2+c2
ac
=
(a+c)2
ac
=
(a+c)2
b(a+c)
2
=
2(a+c)
b

class="stub"b+c
a
class="stub"c+a
b
class="stub"a+b
c
也成等差数列.

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