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
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