2^0+2^1+2^2+...+2^(n-1)为什么用s=(a1-q^(n-1))/(1-q) 算不出正确答案啊
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2^0+2^1+2^2+...+2^(n-1)为什么用s=(a1-q^(n-1))/(1-q) 算不出正确答案啊
2^0+2^1+2^2+...+2^(n-1)
为什么用s=(a1-q^(n-1))/(1-q) 算不出正确答案啊
2^0+2^1+2^2+...+2^(n-1)为什么用s=(a1-q^(n-1))/(1-q) 算不出正确答案啊
你弄错等比数列的前n项和公式了
q≠1时,Sn=a1(1-q^n)/(1-q)中q^n次方中n是指数列所求和项的项数,不是最后一项的次数,
故2^0+2^1+2^2+...+2^(n-1)项数是n项
故
2^0+2^1+2^2+...+2^(n-1)=2^0(1-2^n)/(1-2)=2^n-1.
s=(a1-q^n)/(1-q)
2^n/n*(n+1)
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