已知数列{an}满足a1=0,a2=2,且对任意m'n属于N*,都有a(2m-1)+a(2n+1)=2a(m+n-1)+2(m-n)^2设cn=(a(n+1)-an)q^(n-1),求数列{cn}的前n项和Sn
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![已知数列{an}满足a1=0,a2=2,且对任意m'n属于N*,都有a(2m-1)+a(2n+1)=2a(m+n-1)+2(m-n)^2设cn=(a(n+1)-an)q^(n-1),求数列{cn}的前n项和Sn](/uploads/image/z/5472328-40-8.jpg?t=%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97%7Ban%7D%E6%BB%A1%E8%B6%B3a1%3D0%2Ca2%3D2%2C%E4%B8%94%E5%AF%B9%E4%BB%BB%E6%84%8Fm%27n%E5%B1%9E%E4%BA%8EN%2A%2C%E9%83%BD%E6%9C%89a%282m-1%29%2Ba%282n%2B1%29%3D2a%28m%2Bn-1%29%2B2%28m-n%29%5E2%E8%AE%BEcn%3D%28a%28n%2B1%29-an%29q%5E%28n-1%29%2C%E6%B1%82%E6%95%B0%E5%88%97%7Bcn%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8CSn)
已知数列{an}满足a1=0,a2=2,且对任意m'n属于N*,都有a(2m-1)+a(2n+1)=2a(m+n-1)+2(m-n)^2设cn=(a(n+1)-an)q^(n-1),求数列{cn}的前n项和Sn
已知数列{an}满足a1=0,a2=2,且对任意m'n属于N*,都有a(2m-1)+a(2n+1)=2a(m+n-1)+2(m-n)^2
设cn=(a(n+1)-an)q^(n-1),求数列{cn}的前n项和Sn
已知数列{an}满足a1=0,a2=2,且对任意m'n属于N*,都有a(2m-1)+a(2n+1)=2a(m+n-1)+2(m-n)^2设cn=(a(n+1)-an)q^(n-1),求数列{cn}的前n项和Sn
a(2n-1)+a(2n+1)=2a(n+n-1)+0=2a(2n-1),
a(2n+1)=a(2n-1)=...=a(1)=0.
a(2n-1)=0.
a[2(n+1)-1] + a(2n+1) = 2a(n+1 + n-1) + 2(n+1-n)^2,
a(2n+1) + a(2n+1) = 0 = 2a(2n) + 2,
a(2n)=-1, 与a(2)=2矛盾哈.
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