a(n+1)-a(n)=3n+4
階差数列が3n+4なので、これをb(n)とすると
a(n)=a(1) + Σ[k=1からn-1] b(k)
を解けば出ます。
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a(n+1)-a(n)=3n+4
階差数列が3n+4なので、これをb(n)とすると
a(n)=a(1) + Σ[k=1からn-1] b(k)
を解けば出ます。
Users viewing this question
are also looking at these questions 😉
