数列{an}的前n项和为Sn,a1=a,a(n+1)=Sn+3的n次方,n是整数,a(n+1)大于等于an,求a的取值范

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数列{an}的前n项和为Sn,a1=a,a(n+1)=Sn+3的n次方,n是整数,a(n+1)大于等于an,求a的取值范围

a(n+1)=S(n) + 3^n (1)
a(n) = S(n-1)+ 3^(n-1) (2)
(1)-(2)有
a(n+1) - a(n) = [S(n) + 3^n] - [S(n-1)+ 3^(n-1)] =a(n) + 2*3^(n-1)
a(n+1) = 2a(n) + 2*3^(n-1) >= a(n)
a(n) >= -2*3^(n-1)
其中,a(1) >= -2 可知,a >= -2

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