1.已知两个等差数列An,Bn,前n项和分别为Sn,Tn,且Sn/Tn=(2n+2)/(n+2),则An/Bn=2.已知数列An中,a1+3a2+5a3+...+(2n-1)an=(2n-3)*2^(n+1),求数列的通项公式3.数列a1+2a2+3a3+...+nan=n(n+1)(n+2),求an最后一道不用回答啦 我知
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![1.已知两个等差数列An,Bn,前n项和分别为Sn,Tn,且Sn/Tn=(2n+2)/(n+2),则An/Bn=2.已知数列An中,a1+3a2+5a3+...+(2n-1)an=(2n-3)*2^(n+1),求数列的通项公式3.数列a1+2a2+3a3+...+nan=n(n+1)(n+2),求an最后一道不用回答啦 我知](/uploads/image/z/693409-49-9.jpg?t=1.%E5%B7%B2%E7%9F%A5%E4%B8%A4%E4%B8%AA%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97An%2CBn%2C%E5%89%8Dn%E9%A1%B9%E5%92%8C%E5%88%86%E5%88%AB%E4%B8%BASn%2CTn%2C%E4%B8%94Sn%2FTn%3D%282n%2B2%29%2F%28n%2B2%29%2C%E5%88%99An%2FBn%3D2.%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97An%E4%B8%AD%2Ca1%2B3a2%2B5a3%2B...%2B%282n-1%29an%3D%282n-3%29%2A2%5E%28n%2B1%29%2C%E6%B1%82%E6%95%B0%E5%88%97%E7%9A%84%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F3.%E6%95%B0%E5%88%97a1%2B2a2%2B3a3%2B...%2Bnan%3Dn%28n%2B1%29%28n%2B2%29%2C%E6%B1%82an%E6%9C%80%E5%90%8E%E4%B8%80%E9%81%93%E4%B8%8D%E7%94%A8%E5%9B%9E%E7%AD%94%E5%95%A6+%E6%88%91%E7%9F%A5)
1.已知两个等差数列An,Bn,前n项和分别为Sn,Tn,且Sn/Tn=(2n+2)/(n+2),则An/Bn=2.已知数列An中,a1+3a2+5a3+...+(2n-1)an=(2n-3)*2^(n+1),求数列的通项公式3.数列a1+2a2+3a3+...+nan=n(n+1)(n+2),求an最后一道不用回答啦 我知
1.已知两个等差数列An,Bn,前n项和分别为Sn,Tn,且Sn/Tn=(2n+2)/(n+2),则An/Bn=
2.已知数列An中,a1+3a2+5a3+...+(2n-1)an=(2n-3)*2^(n+1),求数列的通项公式
3.数列a1+2a2+3a3+...+nan=n(n+1)(n+2),求an
最后一道不用回答啦 我知道了
改一道
已知两个等差数列an bn的前n项和为Sn Tn,且Sn/Tn=(7n+2)/(n+3),求a5/b5
三道都要回答哦 好的会加分的
1.已知两个等差数列An,Bn,前n项和分别为Sn,Tn,且Sn/Tn=(2n+2)/(n+2),则An/Bn=2.已知数列An中,a1+3a2+5a3+...+(2n-1)an=(2n-3)*2^(n+1),求数列的通项公式3.数列a1+2a2+3a3+...+nan=n(n+1)(n+2),求an最后一道不用回答啦 我知
1.S2n+1=(A1+A2n+1)*(2n+1)/2=(2n+1)*An(由等差中项推导出来),同理T2n+1=(2n+1)*Bn.所以An/Bn=S2n+1/T2n+1=(4n+4)/(2n+3)
2.由题设a1+3a2+5a3+...+(2n-1)an=(2n-3)*2^(n+1)……①,将1式中的项数换为n-1.有:a1+3a2+5a3+...+(2n-3)an-1=(2n-5)*2^n……②,1式减去2式,得:(2n-1)an=(2n-3)*2^(n+1)-=(2n-5)*2^n=(2n-1)2^n,所以an=2^n
3.和上一道题是一样的,a1+2a2+3a3+...+nan=n(n+1)(n+2)……①,令1式中的项数n为n-1,得:a1+2a2+3a3+...+(n-1)an-1=n(n+1)(n-1),1式减去2式,得:nan=3n(n+1),所以an=3(n+1)