设数列{an}的前n项和为Sn,且（3-m）Sn+2MAn=m+3（m∈N*）,其中m为常数且m≠-3,求证：{an}为等比数列._数学解答

设数列{an}的前n项和为Sn,且（3-m）Sn+2MAn=m+3（m∈N*）,其中m为常数且m≠-3,求证：{an}为等比数列.

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解答

证：n=1时,(3-m)S1+2ma1=(3-m)a1+2ma1=3a1+ma1=m+3a1(m+3)=m+3m≠-3,a1=1n≥2时,(3-m)Sn+2man=m+3 (1)(3-m)S(n-1)+2ma(n-1)=m+3 (2)(1)-(2)(3-m)an+2man-2ma(n-1)=0(m+3)an=2ma(n-1)an/a(n-1)=2m/(m+3),为定值.数列{...