问题
填空题
若等比数列{an}的前n项和Sn满足:an+1=a1Sn+1(n∈N*),则a1=______.
答案
由题意得an+1=a1sn+1 ① an+2=a1sn+1 +1 ②
②-①,得an+1(q-1)=a1an+1,
即a1=q-1,亦即q=1+a1,
所以当n=1时,a2=a1S1+1,
则有a1q=a12+1,即a1(1+a1)=a12+1,
解得a1=1.
故答案为1.
若等比数列{an}的前n项和Sn满足:an+1=a1Sn+1(n∈N*),则a1=______.
由题意得an+1=a1sn+1 ① an+2=a1sn+1 +1 ②
②-①,得an+1(q-1)=a1an+1,
即a1=q-1,亦即q=1+a1,
所以当n=1时,a2=a1S1+1,
则有a1q=a12+1,即a1(1+a1)=a12+1,
解得a1=1.
故答案为1.