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Ϊʲô1/2+1/4+1/6+1/8+.....+1/2n=1?

Ǹ Ӧ1/2+1/4+..+1/2^n=1 ,q=1/2ĵȱ,s=(1/2)/1-(1/2)=11/2+1/4+1/6+1/8Ѿ1

1.

1/2+1/4+1/6+1/8++1/2n+Dz1,Ǵ1.:1/2+1/4+1/6+1/8+.+1/2n+> 1/2+1/4+1/8+1/16+.+1/(2^n)+=1/2+(1/2)^2+(1/2)^3++(1/2)^n+=(1/2)/(1-1/2)=1

Ѷ֮һȡľǸ˹

public classtest { public static void main(string[] args) { double sum = 0; for ( int i = 1; i <= 10; i++){ sum += 1.0 / i; } system.out.println( 1+1/2+1/3+1/4+1/5+1/6+1/7+1/8+1/9+1/10= + sum ); } }

s=1/2+1/4++1/8+1/16+1/2^n2s=1+1/2+1/4+1/9+1/16+1/2^n-12s-s=1-1/2^ns=1-1/2^n1/2+1/4+1/8+1/16+1/2^n=1-1/2^n

ԭʽ=1/2*(1+1/2+1/3+1/4++1/n)

õȱ1/2+(1/2)^2+(1/2)^3++(1/2)^n=1/2*(1-(1/2)^(n-1))/(1-1/2) nʱ(1/2)^(n-1)=0 =1

ȱ (1/2)[1-(1/2)^n]/[1-(1/2)] =1-(1/2)^n

1/2,1/4,1/8,,1/2nη,,,,,1/2Ϊ,1/2Ϊ ĵȱ.ǰnĺ,Ҫ.ûѧ˹ʽ,ԼҲƳ:Sn=1/2+1/4+1/8++1/2nη,__________(1)2Sn=1+1/2+1/4+1/8++1/2(n-1)η,____(2)(2)-(1):Sn=1+(1/2+1/4+1/8++1/2(n-1)η)-(1/2+1/4+1/8++1/2nη) =ȥһ,1-{1/2n}..㿴?

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