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已知数列an满足a1=0,a2=2,且对任意m,n∈N*都有a2m-1+a2n-1=2am+n-1+2(m-n)2 (Ⅰ)求a3,a5; (Ⅱ)设bn=a2n+1-a2n-1(n∈N*),证明:bn是等差数列; (Ⅲ)设cn=(an+1-an)qn-1(q≠0,n∈N*),求数列cn的前n项和Sn.
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