高中数学数列第1问，an的通项公式怎么求，需过程，只求an

由[an+1]=(n+1)/n*an=(n+1)/n*n/(n-1)[an-1]=(n+1)[a1]得[an]=na1;
故-a2=-2a1
a4=4a1
-a5-3=-5a1-3。
故有（4a1）^2=-2a1*(-5a1-3)，解得a1=0或1，舍去0解故有a1=1，an=n。

na(n+1)=(n+1)an

a(n+1)/an = (n+1)/n
an/a(n-1) = n/(n-1)
an/a1 = n
an = a1.n
-a2,a4, -a5-3 是等比数列{bn}的前3项
b1=-a2 = -2a1 (1)
b2=a4= 4a1 (2)
b3=-a5-3 = -5a1 -3 (3)
b1.b3=(b2)^2
(4a1)^2 = (-2a1)(-5a1-3)
16(a1)^2 = 10(a1)^2 +6a1
a1(a1-1)=0
a1=1
=>
an = a1n =n
b1=-a2 = -2
b2 = a4 = 4
b1q=4
q= -2
bn = (-2)^n