已知等比数列{an}的前n项和为Sn，并且对任意正整数n都有Sn+2=4Sn+3...
解：∵等比数列{an}的前n项和为Sn，并且对任意正整数n都有Sn+2=4Sn+3成立，∴Sn+1=4Sn-1+3与Sn+2=4Sn+3两式相减得出：an+2=4an
即an+2an=4，
∵等比数列{an}
∴q2=4，q=±2，
当q=2时，a1×1-2n+21-2=4a1×1-2n1-2+3，a1=1，
∴a2=2，
当q=-2时，a1×1-(-2)n+23=4a1×1-(-2)n3+3，a1=-3，
∴a2=6，
故答案为：2或6
绛旓細绠鍗曞垎鏋愪竴涓嬶紝绛旀濡傚浘鎵绀
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