设等差数列{an}的前n项和为Sn,且S4=4S2,a4=2a2+1,求数列{an}的通项公式及其前n项和Sn
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![设等差数列{an}的前n项和为Sn,且S4=4S2,a4=2a2+1,求数列{an}的通项公式及其前n项和Sn](/uploads/image/z/1132009-25-9.jpg?t=%E8%AE%BE%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%EF%BD%9Ban%EF%BD%9D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BASn%2C%E4%B8%94S4%3D4S2%2Ca4%3D2a2%2B1%2C%E6%B1%82%E6%95%B0%E5%88%97%EF%BD%9Ban%EF%BD%9D%E7%9A%84%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F%E5%8F%8A%E5%85%B6%E5%89%8Dn%E9%A1%B9%E5%92%8CSn)
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设等差数列{an}的前n项和为Sn,且S4=4S2,a4=2a2+1,求数列{an}的通项公式及其前n项和Sn
设等差数列{an}的前n项和为Sn,且S4=4S2,a4=2a2+1,求数列{an}的通项公式及其前n项和Sn
设等差数列{an}的前n项和为Sn,且S4=4S2,a4=2a2+1,求数列{an}的通项公式及其前n项和Sn
由等差数列前n项和公式:sn=na1+n(n-1)d/2,且
s4=4s2 可得 4a1+6d=4(2a1+d) 化简得 2a1=d (1)式
又由a4=2a2+1 可得 a1+3d=2(a1+d)+1 化简得 d-a1=1 (2)式
由(1)(2)两式可解得 a1=1,d=2
于是an=a1+(n-1)d=2n-1 sn=na1+n(n-1)d/2=n+n(n-1)*2=n^2
an = a1+(n-1)d
S4=4S2
(2a1+3d)2 = 4(2a1+d)
2a1=d (1)
a4=2a2+1
a1+3d = 2(a1+d)+1
7a1=6a1+1
a1=1
from (1) d=1/2
an = 1+ (1/2)(n-1) = (1/2)(n+1)
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