已知数列{an}的前n项和Sn=-an-1/2^n-1+2(n为整数2) (1)令bn=2^n×an,求证数列{bn}是等差数列,并求数并求数列{an}的通项公式2)令Cn=(n+1)|n*an,求Tn=c1+c2+...+cn
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![已知数列{an}的前n项和Sn=-an-1/2^n-1+2(n为整数2) (1)令bn=2^n×an,求证数列{bn}是等差数列,并求数并求数列{an}的通项公式2)令Cn=(n+1)|n*an,求Tn=c1+c2+...+cn](/uploads/image/z/2707560-0-0.jpg?t=%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97%7Ban%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8CSn%3D-an-1%2F2%5En-1%2B2%EF%BC%88n%E4%B8%BA%E6%95%B4%E6%95%B02%EF%BC%89+%EF%BC%881%EF%BC%89%E4%BB%A4bn%3D2%5En%C3%97an%2C%E6%B1%82%E8%AF%81%E6%95%B0%E5%88%97%7Bbn%7D%E6%98%AF%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%2C%E5%B9%B6%E6%B1%82%E6%95%B0%E5%B9%B6%E6%B1%82%E6%95%B0%E5%88%97%7Ban%7D%E7%9A%84%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F2%EF%BC%89%E4%BB%A4Cn%3D%28n%2B1%29%7Cn%2Aan%2C%E6%B1%82Tn%3Dc1%2Bc2%2B...%2Bcn)
已知数列{an}的前n项和Sn=-an-1/2^n-1+2(n为整数2) (1)令bn=2^n×an,求证数列{bn}是等差数列,并求数并求数列{an}的通项公式2)令Cn=(n+1)|n*an,求Tn=c1+c2+...+cn
已知数列{an}的前n项和Sn=-an-1/2^n-1+2(n为整数2) (1)令bn=2^n×an,求证数列{bn}是等差数列,并求数
并求数列{an}的通项公式
2)令Cn=(n+1)|n*an,求Tn=c1+c2+...+cn
已知数列{an}的前n项和Sn=-an-1/2^n-1+2(n为整数2) (1)令bn=2^n×an,求证数列{bn}是等差数列,并求数并求数列{an}的通项公式2)令Cn=(n+1)|n*an,求Tn=c1+c2+...+cn
Sn=-an-1/2^n-1+2(n>=2).1
Sn-1=-a(n-1)-1/2^(n-2)+2.2
1-2得:an=an-1-an-1/2(n-2)
an=a(n-1)/2-1/2(n-1)
上式左右同乘以2^n得
2^nan=2^(n-1)a(n-1)-2
即bn=b(n-1)-2
即bn为等差数列.
(1)在Sn=-an-(12)n-1+2中令n=1可得s1=-a1-1+2=a1即a1=12
当n≥2时an=Sn-Sn-1=-an+an-1+(
12)n-1
∴2an=an-1+(
12)n-1即2nan=2n-1an-1+1
∵bn=2nan,
∴bn-bn-1=1即当n≥2时bn-bn-1=1
又∵b1=2a1=1
∴数列{b...
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(1)在Sn=-an-(12)n-1+2中令n=1可得s1=-a1-1+2=a1即a1=12
当n≥2时an=Sn-Sn-1=-an+an-1+(
12)n-1
∴2an=an-1+(
12)n-1即2nan=2n-1an-1+1
∵bn=2nan,
∴bn-bn-1=1即当n≥2时bn-bn-1=1
又∵b1=2a1=1
∴数列{bn}是首项和公差均为1的等差数列.
∴bn=1+(n-1)×1=n=2nan
∴an=
n2n
(2)由(1)得cn=(n+1)(
12)n,
∴Tn=2×
12+3×(
12)2+4×(
12)3+…+(n+1)(
12)n ①
12Tn=2×(
12)2+3×(
12)3+4×(
12)4+…+(n+1)(
12)n+1 ②
由①-②得12Tn=1+(
12)2+(
12)3+…+(
12)n-(n+1)(
12)n+1=32-n+32n
∴Tn=3-n+32n
收起
an=n*(0.5)^n
bn=n