Question

在数列{an}中,已知an=2,对任意正整数n都有nan+1=2(n+1)an,求an的通项公式

Answer 1

na(n+1)=2(n+1)an
a(n+1)/an = 2(n+1)/n
an/a(n-1) = 2n/(n-1)
an/a1 = 2^(n-1) . n
an = n.2^n

追问

我要详细答案啊

追答

na(n+1)=2(n+1)an
a(n+1)/an = 2(n+1)/n
an/a(n-1) = 2n/(n-1) (1)
a(n-1)/a(n-2) = 2(n-1)/(n-2) (2)
..
...
a2/a1 = 2(2)/1 (n-1)
(1)*(2)*(3)*...*(n-1)
an/a1 = 2^(n-1). n/1
an/a1 = 2^(n-1) . n
an = n.2^n

Answer 2

n.a(n+1) =2(n+1)an
a(n+1)/an = 2(n+1)/n
an/a(n-1) = 2n/(n-1)
an/a1 = n 2^(n-1)
an = n.2^n

追问

详细答案啊

追答

这已经很详细了啊？你是哪部看不懂么？

追问

an/a1 = n 2^(n-1)这里怎么来？

追答

这里是前一个等式n-1项相乘的结果，中间大部分都约掉了，只剩an/a1，而右边除了n/1之外，还有2的累乘，由于有n-1项，所以是2^(n-1)