Question

第七题怎么写

Answer 1

由an+1+an−1an+1−an+1=n可得an+1+an−1=nan+1−nan+n
∴(1−n)an+1+(1+n)an=1+n
∴an+1=n+1n−1an−n+1n−1=1n−1(an−1)×(n+1)
∴an+1n+1=1n−1an−1n−1=nn−1⋅ann−nn−1⋅1n
∴an+1n+1−1=nn−1(ann−1)
∴1n(an+1n+1−1)=1n−1(ann−1)
∴{1n−1(ann−1)}为常数列
而1n−1(ann−1)=12−1⋅(a22−1)=2，
an=[2(n−1)+1]n=2n2−n
当n=1时，6+a1−16−a1+1=1可得a1=1适合上式
故：2n2−n

追问

请问第五步怎么会有个➖1

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