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答:
{an}前2n项中有n个奇数项和n个偶数项,设奇数项构成数列{bn},偶数项构成数列{cn}
则b1=a1=1,{bn}公比为a3/a1=2,所以bn=2^(n-1)
c1=a2=1/2,{cn}公比为a4/a2=1/2,所以cn=1/2^n
记{bn}前n项和为Tn,{cn}前n项和为Un,则:
S2n=Tn+Un
=1×(1-2^n)/(1-2)+1/2×(1-1/2^n)/(1-1/2)
=2^n-1+1-1/2^n
=2^n-1/2^n
1年前
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