已知数列{an}满足a1=1;an=a1+2a2+3a3+...+(n-1)a(n-1);(n>=2);求通项公式.(请给出过程,谢谢)
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![已知数列{an}满足a1=1;an=a1+2a2+3a3+...+(n-1)a(n-1);(n>=2);求通项公式.(请给出过程,谢谢)](/uploads/image/z/7096050-18-0.jpg?t=%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97%7Ban%7D%E6%BB%A1%E8%B6%B3a1%3D1%3Ban%3Da1%2B2a2%2B3a3%2B...%2B%28n-1%29a%28n-1%29%3B%EF%BC%88n%3E%3D2%29%3B%E6%B1%82%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F.%EF%BC%88%E8%AF%B7%E7%BB%99%E5%87%BA%E8%BF%87%E7%A8%8B%2C%E8%B0%A2%E8%B0%A2%EF%BC%89)
已知数列{an}满足a1=1;an=a1+2a2+3a3+...+(n-1)a(n-1);(n>=2);求通项公式.(请给出过程,谢谢)
已知数列{an}满足a1=1;an=a1+2a2+3a3+...+(n-1)a(n-1);
(n>=2);求通项公式.(请给出过程,谢谢)
已知数列{an}满足a1=1;an=a1+2a2+3a3+...+(n-1)a(n-1);(n>=2);求通项公式.(请给出过程,谢谢)
a2=a1+2a2=1+2a2 得a2=-1
an=a1+2a2+3a3+...+(n-2)a(n-2)+(n-1)a(n-1)
a(n-1)=a1+2a2+3a3+...+(n-2)a(n-2)
两式相减:
an-a(n-1)=(n-1)a(n-1)
即an=na(n-1)
所以an/a(n-1)=n
an=[an/a(n-1)][a(n-1)/a(n-2)]...[a3/a2]a2=
n*(n-1)*...*3a2=(-1)n!/2
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