若等比数列an的前n项和为sn=2^n+a , 求a

令n=1,得S1=a1=2+a
令n=2,得S2=a1+a2=4+a
两式相减,得a2=2
同样,令n=3,得S3=a1+a2+a3=8+a
所以a3=4

这是一个等比数列,所以公比=a3/a2=2
所以a1=a2/公比=1
2+a=a1
所以a=-1

很久没做过数学了

还有什么不懂的可以问我

s(n) = 2^n + a, n = 1,2,...

s(n+1) = 2^(n+1) + a,

a(n+1) = s(n+1) - s(n) = 2^(n+1) - 2^n = 2^(n+1)[1 - 1/2] = 2^(n+1)/2,

所以，a(n) = 2^n/2 = 2^(n-1), n = 1,2,...

2 + a = s(1) = a(1) = 2^(1-1) = 1, a = -1

an=sn-s（n-1）=2^(n-1)a1=s1=1=2+a
a=-1

s(n) = 2^n + a, n = 1,2,...

s(n+1) = 2^(n+1) + a
an = s(n+1) - s(n) = 2^(n+1) - 2^n = 2^n[1 - 1/2] = 2^n/2,
以，a(n) = 2^n/2 = 2^(n-1), n = 1,2,...

2 + a = s(1) = a(1) = 2^(1-1) = 1, a = -1