等比数列{an}的前n项和Sn=2*3^(n+1)+k,首项a1=

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等比数列{an}的前n项和Sn=2*3^(n+1)+k,首项a1=

等比数列{an}的前n项和Sn=2*3^(n+1)+k,首项a1=
等比数列{an}的前n项和Sn=2*3^(n+1)+k,首项a1=

等比数列{an}的前n项和Sn=2*3^(n+1)+k,首项a1=
sn=a1(1-q^n)/(1-q)=a1(q^n-1)/(q-1)= (a1*q^n-a1)/(q-1) = 2*3^(n+1)+k = 6*3^n+k
所以q=3
所以sn = (a1*3^n - a1)/2 = 6*3^n+k
所以a1/2 = 6,a1 = 12
k = -12/2 = -6