已知数列{An}是首项a1=1的等比数列,且An〉0,{bn}是首项为1的等差数列,又a5+b3=21,a3+b5=13 〔1〕求数列...已知数列{An}是首项a1=1的等比数列,且An〉0,{bn}是首项为1的等差数列,又a5+b3=21,a3+b5=13〔1〕求数
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/19 17:19:23
![已知数列{An}是首项a1=1的等比数列,且An〉0,{bn}是首项为1的等差数列,又a5+b3=21,a3+b5=13 〔1〕求数列...已知数列{An}是首项a1=1的等比数列,且An〉0,{bn}是首项为1的等差数列,又a5+b3=21,a3+b5=13〔1〕求数](/uploads/image/z/5448821-5-1.jpg?t=%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97%7BAn%7D%E6%98%AF%E9%A6%96%E9%A1%B9a1%3D1%E7%9A%84%E7%AD%89%E6%AF%94%E6%95%B0%E5%88%97%2C%E4%B8%94An%E3%80%890%2C%7Bbn%7D%E6%98%AF%E9%A6%96%E9%A1%B9%E4%B8%BA1%E7%9A%84%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%2C%E5%8F%88a5%2Bb3%3D21%2Ca3%2Bb5%3D13+%E3%80%941%E3%80%95%E6%B1%82%E6%95%B0%E5%88%97...%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97%7BAn%7D%E6%98%AF%E9%A6%96%E9%A1%B9a1%3D1%E7%9A%84%E7%AD%89%E6%AF%94%E6%95%B0%E5%88%97%2C%E4%B8%94An%E3%80%890%2C%7Bbn%7D%E6%98%AF%E9%A6%96%E9%A1%B9%E4%B8%BA1%E7%9A%84%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%2C%E5%8F%88a5%2Bb3%3D21%2Ca3%2Bb5%3D13%E3%80%941%E3%80%95%E6%B1%82%E6%95%B0)
已知数列{An}是首项a1=1的等比数列,且An〉0,{bn}是首项为1的等差数列,又a5+b3=21,a3+b5=13 〔1〕求数列...已知数列{An}是首项a1=1的等比数列,且An〉0,{bn}是首项为1的等差数列,又a5+b3=21,a3+b5=13〔1〕求数
已知数列{An}是首项a1=1的等比数列,且An〉0,{bn}是首项为1的等差数列,又a5+b3=21,a3+b5=13 〔1〕求数列...
已知数列{An}是首项a1=1的等比数列,且An〉0,{bn}是首项为1的等差数列,又a5+b3=21,a3+b5=13
〔1〕求数列{An}和{Bn}的通项公式;
这问我算出来了,bn=2n-1 an=2^n
〔2〕求{Bn/2An}的前n项和Sn
不好意思,an=2^(n-1)。
已知数列{An}是首项a1=1的等比数列,且An〉0,{bn}是首项为1的等差数列,又a5+b3=21,a3+b5=13 〔1〕求数列...已知数列{An}是首项a1=1的等比数列,且An〉0,{bn}是首项为1的等差数列,又a5+b3=21,a3+b5=13〔1〕求数
这么写就清楚多了~
Sn= 1/2+3/2^2+5/2^3+.+(2n-3)/2^(n-1)+(2n-1)/2^n ①
2Sn=1+3/2+5/2^2+7/2^3+.+(2n-1)/2^(n-1) ②
②-①得
Sn=1+2/2+2/2^2+2/2^3+.+2/2^(n-1)-(2n-1)/2^n
=1+ 1+1/2 +1/2^2+.+1/2^(n-2)-(2n-1)/2^n
=1+1*[1-1/2^(n-1)]/(1-1/2)-(2n-1)/2^n
=1+[2-1/2^(n-2)]-(2n-1)/2^n
=3-4/2^n-(2n-1)/2^n
=3-(2n+3)/2^n
Sn=1/2²+3/2³+......2n-1/2^(n+1) ①
2Sn=1/2+3/2²+.....2n-1/2^n ②
②-①得
Sn=1/2+2(1/2²+1/2³+.......1/2^n)-(2n-1)/2^(n+1)
=1/2+2×[1/2²×(1-0.5^(n-1)/(1-0.5)]-(2n-1)/2^(n+1)
=1/2+1-0.5^(n-1)-(2n-1)/2^(n+1)
=3/2-0.5^(n-1)-(2n-1)/2^(n+1)
Sn=1/4+3/8+...+(2n-3)/2^n+(2n-1)/2^(n+1);
2Sn=1/2+3/4+...+(2n-3)2^(n-1)+(2n-1)/2^n;
Sn=2Sn-Sn=1/2+1/2+1/4...+1/2^(n-1)-(2n-1)/2^(n+1)
=1.5-(2n+3)/2^(n+1)