已知等比数列{an} 满足Sn=a+2^(n-1) 求实数a
人气:186 ℃ 时间:2020-06-18 03:42:35
解答
Sn=[a1(1-q^n]/﹙q-1﹚
有Sn=a+2^(n-1) 可看出没有分母,
则q-1=1,q=2
可得Sn=a1(1-2)^n=a1-a1x2^n
得a1=-1/2
∴Sn=-1/2+2^(n-1)
∴a=-1/2应该是Sn=[a1(1-q^n]/﹙1-q﹚吧? 不是q-1啊抱歉,是1-q1-q=-1,q=2下面同求!
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