等差数列{an}与{bn},它们的前n项之和分别为Sn,S'n,如果Sn/S'n=(7n+1)/(4n+27)(n∈N*),则a11/b11的值是
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![等差数列{an}与{bn},它们的前n项之和分别为Sn,S'n,如果Sn/S'n=(7n+1)/(4n+27)(n∈N*),则a11/b11的值是](/uploads/image/z/1761510-30-0.jpg?t=%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%7Ban%7D%E4%B8%8E%7Bbn%7D%2C%E5%AE%83%E4%BB%AC%E7%9A%84%E5%89%8Dn%E9%A1%B9%E4%B9%8B%E5%92%8C%E5%88%86%E5%88%AB%E4%B8%BASn%2CS%27n%2C%E5%A6%82%E6%9E%9CSn%2FS%27n%3D%287n%2B1%29%2F%284n%2B27%29%28n%E2%88%88N%2A%29%2C%E5%88%99a11%2Fb11%E7%9A%84%E5%80%BC%E6%98%AF)
等差数列{an}与{bn},它们的前n项之和分别为Sn,S'n,如果Sn/S'n=(7n+1)/(4n+27)(n∈N*),则a11/b11的值是
等差数列{an}与{bn},它们的前n项之和分别为Sn,S'n,如果Sn/S'n=(7n+1)/(4n+27)(n∈N*),则a11/b11的值是
等差数列{an}与{bn},它们的前n项之和分别为Sn,S'n,如果Sn/S'n=(7n+1)/(4n+27)(n∈N*),则a11/b11的值是
S(2n-1)=(a1+a2n-1)(2n-1)/2=an(2n-1)
S'(2n-1)=(b1+b2n-1)(2n-1)/2=bn(2n-1)
两式相除:S(2n-1)/S'(2n-1)=an/bn
a11/b11的值=S21/S'21=(7*21+1)/(4*21+27)=4/3
请证明:若数列{n}与{bn}都是等差数列,它们的前n项和分别为Sn,Tn,则an/bn=S2n-1/T2n-1
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