1+1/2+1/4+1/8+1/16+1/32+1/64+1/128+1/256+1/512+1/1024求解

1+1/2+1/4+1/8+1/16+1/32+1/64+1/128+1/256+1/512+1/1024求解
1+1/2+1/4+1/8+1/16+1/32+1/64+1/128+1/256+1/512+1/1024求解

1+1/2+1/4+1/8+1/16+1/32+1/64+1/128+1/256+1/512+1/1024求解
原式=1+1/2+1/4…+1/1024+1/1024-1/1024=2-1/1024=2047/1024.在公比为2的数列中,可加一个最小项,再减去它,答案是两倍最大项减一个最小项,如1/1024+1/1024=1/512,1/512+1/512=1/256……1/2+1/2=1,1+1=2,最后减去一开始加的1/1024.

这是一个等比数列，首项a1=1，公比=1/2，

设为x则有x=1/2(1+1/2...1/1024)+1-1/2048=1/2x+2047/2048
x=2047/1024

令S=1+1/2+1/4+1/8+1/16+1/32+1/64+1/128+1/256+1/512+1/1024 ①
则2S=2+1+1/2+1/4+1/8+1/16+1/32+1/64+1/128+1/256+1/512 ②
②-①得S=2-1/1024=2047/1024

1+(1-1024)=一又一千零二十四分之一千零二十三