等差数列问题,疑惑,差数列an bn的前n项和等差数列an bn的前n项和是S T S/T=2n/(3n+1)求an/bnSn=(n/2)(a1+an)=(n/2)(Dn+2a1-D)……设D是数列{an}的公差Tn=(n/2)(b1+bn)=(n/2)(dn+2b1-d)……设d是数列{bn}的公差Sn/Tn=(Dn+2a1
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![等差数列问题,疑惑,差数列an bn的前n项和等差数列an bn的前n项和是S T S/T=2n/(3n+1)求an/bnSn=(n/2)(a1+an)=(n/2)(Dn+2a1-D)……设D是数列{an}的公差Tn=(n/2)(b1+bn)=(n/2)(dn+2b1-d)……设d是数列{bn}的公差Sn/Tn=(Dn+2a1](/uploads/image/z/10926766-46-6.jpg?t=%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%E9%97%AE%E9%A2%98%2C%E7%96%91%E6%83%91%2C%E5%B7%AE%E6%95%B0%E5%88%97an+bn%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97an+bn%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E6%98%AFS+T+S%2FT%3D2n%2F%283n%2B1%29%E6%B1%82an%2FbnSn%3D%28n%2F2%29%28a1%2Ban%29%3D%28n%2F2%29%28Dn%2B2a1-D%29%E2%80%A6%E2%80%A6%E8%AE%BED%E6%98%AF%E6%95%B0%E5%88%97%7Ban%7D%E7%9A%84%E5%85%AC%E5%B7%AETn%3D%28n%2F2%29%28b1%2Bbn%29%3D%28n%2F2%29%28dn%2B2b1-d%29%E2%80%A6%E2%80%A6%E8%AE%BEd%E6%98%AF%E6%95%B0%E5%88%97%7Bbn%7D%E7%9A%84%E5%85%AC%E5%B7%AESn%2FTn%3D%28Dn%2B2a1)
等差数列问题,疑惑,差数列an bn的前n项和等差数列an bn的前n项和是S T S/T=2n/(3n+1)求an/bnSn=(n/2)(a1+an)=(n/2)(Dn+2a1-D)……设D是数列{an}的公差Tn=(n/2)(b1+bn)=(n/2)(dn+2b1-d)……设d是数列{bn}的公差Sn/Tn=(Dn+2a1
等差数列问题,疑惑,差数列an bn的前n项和
等差数列an bn的前n项和是S T
S/T=2n/(3n+1)求an/bn
Sn=(n/2)(a1+an)=(n/2)(Dn+2a1-D)……设D是数列{an}的公差
Tn=(n/2)(b1+bn)=(n/2)(dn+2b1-d)……设d是数列{bn}的公差
Sn/Tn=(Dn+2a1-D)/(dn+2b1-d)=2n/(3n+1)
所以 设D=2k,则 2a1-D=0,d=3k,2b1-d=k
所以 a1=k,D=2k
b1=2k,d=3k
an=k+(n-1)·2k
bn=2k+(n-1)·3k
为什么?
设D=2k,则 2a1-D=0,d=3k,2b1-d=k
等差数列问题,疑惑,差数列an bn的前n项和等差数列an bn的前n项和是S T S/T=2n/(3n+1)求an/bnSn=(n/2)(a1+an)=(n/2)(Dn+2a1-D)……设D是数列{an}的公差Tn=(n/2)(b1+bn)=(n/2)(dn+2b1-d)……设d是数列{bn}的公差Sn/Tn=(Dn+2a1
2n/(3n+1)是最简分数,其实它应该是2nk/(3nk+k)的约分,k是不为零实数.
2n/(3n+1)=2nk/(3nk+k)
那么(Dn+2a1-D)/(dn+2b1-d)=2n/(3n+1)
即(Dn+2a1-D)/(dn+2b1-d)=2nk/(3nk+k),对照系数就知道
设D=2k,则 2a1-D=0,d=3k,2b1-d=k
Sn=(n/2)(a1+an)=(n/2)(Dn+2a1-D)……设D是数列{an}的公差
Tn=(n/2)(b1+bn)=(n/2)(dn+2b1-d)……设d是数列{bn}的公差
Sn/Tn=(D*n+2a1-D)/(dn+2b1-d)=2n/(3n+1)
(D*n+2a1-D)项2a1-D是常数与n无关,不随n变化,因此2a1-D=0
D*n=2n D是偶数...
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Sn=(n/2)(a1+an)=(n/2)(Dn+2a1-D)……设D是数列{an}的公差
Tn=(n/2)(b1+bn)=(n/2)(dn+2b1-d)……设d是数列{bn}的公差
Sn/Tn=(D*n+2a1-D)/(dn+2b1-d)=2n/(3n+1)
(D*n+2a1-D)项2a1-D是常数与n无关,不随n变化,因此2a1-D=0
D*n=2n D是偶数,因此D=2k
所以d*n+2b1-d=3kn+k
2b1-d是常数,和n无关,不随n变化,2b1-d=k
因此d=3k
所以 设D=2k, 则 2a1-D=0, d=3k, 2b1-d=k
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