已知数列{an}满足a1=0,a2=2,且对任意m、n∈N都有a(2m-1)+a(2n-1)=2a(m+n-1)+2(m-n)^2(1)求a3,a5(2)设bn=s(2n+1)-a(2n-1)(n∈N+),证明:{bn}是等差数列(3)设cn=(a(n+1)-an)q^(n-1)(q≠0,n∈N+),求数列{cn}的前n项和sn以上、
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![已知数列{an}满足a1=0,a2=2,且对任意m、n∈N都有a(2m-1)+a(2n-1)=2a(m+n-1)+2(m-n)^2(1)求a3,a5(2)设bn=s(2n+1)-a(2n-1)(n∈N+),证明:{bn}是等差数列(3)设cn=(a(n+1)-an)q^(n-1)(q≠0,n∈N+),求数列{cn}的前n项和sn以上、](/uploads/image/z/5531066-26-6.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%E3%80%81n%E2%88%88N%E9%83%BD%E6%9C%89a%282m-1%29%2Ba%282n-1%29%3D2a%28m%2Bn-1%29%2B2%28m-n%29%5E2%281%29%E6%B1%82a3%2Ca5%282%29%E8%AE%BEbn%3Ds%282n%2B1%29-a%282n-1%29%28n%E2%88%88N%2B%29%2C%E8%AF%81%E6%98%8E%EF%BC%9A%7Bbn%7D%E6%98%AF%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%283%29%E8%AE%BEcn%3D%28a%28n%2B1%29-an%29q%5E%28n-1%29%28q%E2%89%A00%2Cn%E2%88%88N%2B%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%E4%BB%A5%E4%B8%8A%E3%80%81)
已知数列{an}满足a1=0,a2=2,且对任意m、n∈N都有a(2m-1)+a(2n-1)=2a(m+n-1)+2(m-n)^2(1)求a3,a5(2)设bn=s(2n+1)-a(2n-1)(n∈N+),证明:{bn}是等差数列(3)设cn=(a(n+1)-an)q^(n-1)(q≠0,n∈N+),求数列{cn}的前n项和sn以上、
已知数列{an}满足a1=0,a2=2,且对任意m、n∈N都有a(2m-1)+a(2n-1)=2a(m+n-1)+2(m-n)^2
(1)求a3,a5
(2)设bn=s(2n+1)-a(2n-1)(n∈N+),证明:{bn}是等差数列
(3)设cn=(a(n+1)-an)q^(n-1)(q≠0,n∈N+),求数列{cn}的前n项和sn
以上、
已知数列{an}满足a1=0,a2=2,且对任意m、n∈N都有a(2m-1)+a(2n-1)=2a(m+n-1)+2(m-n)^2(1)求a3,a5(2)设bn=s(2n+1)-a(2n-1)(n∈N+),证明:{bn}是等差数列(3)设cn=(a(n+1)-an)q^(n-1)(q≠0,n∈N+),求数列{cn}的前n项和sn以上、
给点鼓励吧,多谢了!
1,当m=2,n=2时,a(2x2-1)+a(2x2-1)=2a(2+2-1)+2(2-2)^2
即,a3=0
当m=3,n=3时,a(2x3-1)+a(2x3-1)=2a(3+3-1)+2(3-3)^2
即,a5=0
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