已知An的前n项和为Sn且Sn=2n²+n,bn满足An=4log2bn+3. 求⑴An
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解决时间 2021-11-29 09:21
- 提问者网友：活着好累
- 2021-11-28 21:51
已知An的前n项和为Sn且Sn=2n²+n,bn满足An=4log2bn+3. 求⑴An
最佳答案
- 五星知识达人网友：罪歌
- 2021-11-28 23:15
Sn=2n^2+n
n=1, a1=3
for n>=2
an = Sn - S(n-1)
=2(2n-1) +1
=4n -1
an = 4log<2>bn +3
log<2>bn = n- 1
bn = 2^(n-1)
let
S =1.2^1+2.2^2+...+n.2^n (1)
2S = 1.2^2+2.2^3+...+n.2^(n+1) (2)
(2)-(1)
S = n.2^(n+1)-(2+2^2+...+2^n)
=n.2^(n+1)-2(2^n -1)
cn =an.bn
=(4n-1).2^(n-1)
= 2(n.2^n) - 2^(n-1)
Tn =c1+c2+...+cn
=2S - (2^n -1)
=2n.2^(n+1)-5(2^n -1)
= 5 + (4n-5).2^n
n=1, a1=3
for n>=2
an = Sn - S(n-1)
=2(2n-1) +1
=4n -1
an = 4log<2>bn +3
log<2>bn = n- 1
bn = 2^(n-1)
let
S =1.2^1+2.2^2+...+n.2^n (1)
2S = 1.2^2+2.2^3+...+n.2^(n+1) (2)
(2)-(1)
S = n.2^(n+1)-(2+2^2+...+2^n)
=n.2^(n+1)-2(2^n -1)
cn =an.bn
=(4n-1).2^(n-1)
= 2(n.2^n) - 2^(n-1)
Tn =c1+c2+...+cn
=2S - (2^n -1)
=2n.2^(n+1)-5(2^n -1)
= 5 + (4n-5).2^n
全部回答
- 1楼网友：独行浪子会拥风
- 2021-11-28 23:50
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