Question

等比数列帮忙做下谢谢啦

Answer 1

∵an² + 2an=4Sn + 3
∴a²(n+1) + 2a(n+1)=4S(n+1) + 3
两式相减，得：
a²(n+1) + 2a(n+1) - an² - 2an=4a(n+1)
a²(n+1) - an² - 2a(n+1) - 2an=0
[a(n+1) + an][a(n+1) - an] - 2[a(n+1) + an]=0
[a(n+1) + an][a(n+1) - an - 2]=0
∵an>0
∴a(n+1) + an≠0
∴a(n+1) - an - 2=0
则a(n+1) - an=2
∴数列{an}是等差数列
∵a1=S1
∴a1² + 2a1=4a1 + 3
a1² - 2a1 - 3=0
(a1 - 3)(a1 + 1)=0
∴a1=3或a1=-1(舍)
∴an=a1 + (n-1)d=3 + 2(n-1)=2n+1