作业帮一道基本数列问题
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一道基本数列问题
一道基本数列问题
一道基本数列问题
本题不是等差数列,而你当成了等差数列在求解,就错了.
∵ a(n+₁) = an + 3n + 2
∴ an = a(n-₁) + 3(n-1) + 2
= a(n-₁) + [3(n-1) + 2]
= [a(n-₂)+3(n-2)+2] + [3(n-1)+2]
= a(n-₂)+ [3(n-2)+ 3(n-1)+2×2]
= [a(n-₃)+3(n-3)+2] + [3(n-2)+3(n-1)+2×2]
= a(n-₃)+ [3(n-3)+3(n-2)+3(n-1)+3×2]
= a(n-₄)+ [3(n-4)+3(n-3)+3(n-2)+3(n-1)+4×2]
= a(n-(n-₁))+ [3(n-(n-1))+...+3(n-4)+3(n-3)+3(n-2)+3(n-1)+(n-1)×2]
= a₁+ 3[1+2+3+4+...+(n-1)] + 2(n-1)
= 2 + 3×[(1+n-1)(n-1)/2] + 2(n-1)
= 2 + 3n(n-1)/2 + 2(n-1)
= 3(n-1)n/2 + 2n
= [n/2][3(n-1) + 4n]
= n(7n-3)/2