aboveyusheng
幼苗
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确定sn不是2^n -1?
Sn=2^n -1,
S(n-1)=2^(n-1) -1,
an=Sn-S(n-1)
=2^n -1-[2^(n-1) -1]
=2^n-2^(n-1)
=2^(n-1)
an^2=2^(2n-2)=4^n/4,
a(n+1)^2=4^(n+1)/4,
a(n+1)^2/an^2=4
an^2是以a1^2=1为首项,4为公比的等比数列;
S=(1-4^n)/(1-4)
=(4^n-1)/3
=4^n/3-1/3
1年前
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