求数列数列{an}满足a1=1,且n≥2时,an=a1+2a2+3a3+•••+(n-1)an-1的通项公式求数列{an}满足a1=1,且n≥2时,an=a1+2a2+3a3+•••+(n-1)an-1的通项公式
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求数列数列{an}满足a1=1,且n≥2时,an=a1+2a2+3a3+•••+(n-1)an-1的通项公式求数列{an}满足a1=1,且n≥2时,an=a1+2a
求数列数列{an}满足a1=1,且n≥2时,an=a1+2a2+3a3+•••+(n-1)an-1的通项公式求数列{an}满足a1=1,且n≥2时,an=a1+2a2+3a3+•••+(n-1)an-1的通项公式
求数列数列{an}满足a1=1,且n≥2时,an=a1+2a2+3a3+•••+(n-1)an-1的通项公式
求数列{an}满足a1=1,且n≥2时,an=a1+2a2+3a3+•••+(n-1)an-1的通项公式
求数列数列{an}满足a1=1,且n≥2时,an=a1+2a2+3a3+•••+(n-1)an-1的通项公式求数列{an}满足a1=1,且n≥2时,an=a1+2a2+3a3+•••+(n-1)an-1的通项公式
an=a1+2a2+3a3+•••+(n-1)a_{n-1}
a_{n-1}=a1+2a2+3a3+•••+(n-2)a_{n-2}
两式相减,
a_n-a_{n-1}=(n-1)a_{n-1}
以及 a_2=a_1=1
所以a_n=na_{n-1}
因此 a_n=n!/2 这里n!就是n的阶乘.
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