lim(2+4+6+...+2^n)/(1+3+5+...+2^n-1)lim(1+1/2+1/4+...+1/2^n-1)/(1+1/3+1/9+...+1/3^n-1)怎么解n-无穷大
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lim(2+4+6+...+2^n)/(1+3+5+...+2^n-1)
lim(1+1/2+1/4+...+1/2^n-1)/(1+1/3+1/9+...+1/3^n-1)
怎么解
n-无穷大
2+4+6+...+2^n=n(2+2^n)/2=n[1+2^(n-1)]
1+3+5+...+(2^n-1)=(2+4+6+...+2^n)-n=n*2^(n-1)
lim(2+4+6+...+2^n)/(1+3+5+...+2^n-1)
=lim{n[1+2^(n-1)]}/[n*2^(n-1)]
=lim[1+2^(n-1)]/2^(n-1)
=lim[1/2^(n-1)+1]
=1
1+1/2+1/4+...+1/2^(n-1)=2-1/2^(n-1)
1+1/3+1/9+...+1/3^(n-1)=1/2[3-1/3^(n-1)]
lim[1+1/2+1/4+...+1/2^(n-1)]/[1+1/3+1/9+...+1/3^(n-1)]
=lim2[2-1/2^(n-1)]/[3-1/3^(n-1)]
=2lim2/3
=4/3
分子=(1-(1/2)^n+1)/(1/2)-1 =1-(1/2)^n ->1
分母 =(1-(1/3)^n+1)/(2/3)-1 ->1/2
所以极限是2
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