2/1+4/1+8/1

8-1/2×4/16+1/8?8-1/2×4/16+1/8?8-1/2×4/16+1/8?31/4

数学题；1-1/2-1/4-1/8-...-1/128数学题；1-1/2-1/4-1/8-...-1/128数学题；1-1/2-1/4-1/8-...-1/1281-1/2-1/4-1/8-...-1

2\1+4\1+8\1+16\1+32\1?2\1+4\1+8\1+16\1+32\1?2\1+4\1+8\1+16\1+32\1?(1-1/2)+(1/2-1/4)+(1/4-1/8)+(1/8-1

1/2+(-1/4)+(-1/8)+(-1/16)+(-1/32)1/2+(-1/4)+(-1/8)+(-1/16)+(-1/32)1/2+(-1/4)+(-1/8)+(-1/16)+(-1/32)1

1/2+1/4+1/8+1/16+...+1/10241/2+1/4+1/8+1/16+...+1/10241/2+1/4+1/8+1/16+...+1/10241/2+1/4+1/8+1/16+..

1+1/2+1/4+1/8+...1/64=?1+1/2+1/4+1/8+...1/64=?1+1/2+1/4+1/8+...1/64=?就是一个等比数列的求和问题.Sn=a1(1-q^n)/(1-q

1+1/2+1/4+1/8+.1/10241+1/2+1/4+1/8+.1/10241+1/2+1/4+1/8+.1/10241+1/2+1/4+1/8+...+1/1024=1+1/2*(1-1/2

1/2+1/4+1/6+1/8+.1/n1/2+1/4+1/6+1/8+.1/n1/2+1/4+1/6+1/8+.1/n这是无穷级数中很有名的调和级数,本身是发散的,极限不存在,也就是说没有定值.

1/2+1/4+1/8+1/16+...+1/n1/2+1/4+1/8+1/16+...+1/n1/2+1/4+1/8+1/16+...+1/n令t=1/2+1/4+1/8+1/16+...+1/2^

1/2+1/4+1/8+...+1/2n1/2+1/4+1/8+...+1/2n1/2+1/4+1/8+...+1/2n原式可看成是首项为1/2,公比为1/2的等比数列的前n项和,则可利用等比数列求和

1/2+1/4+1/8+...+1/2n=?1/2+1/4+1/8+...+1/2n=?1/2+1/4+1/8+...+1/2n=?等比数列求和(1/2)[1-(1/2)^n]/[1-(1/2)]=1

|1/4-1/2|+|1/6-1/4|+|1/8-1/6|+```+|1/2008-1/2006||1/4-1/2|+|1/6-1/4|+|1/8-1/6|+```+|1/2008-1/2006||1

[(1+2^-(1/32)]*[(1+2^-(1/16)]*[(1+2^-(1/8)]*[(1+2^-(1/4)]*[(1+2^-(1/2)][(1+2^-(1/32)]*[(1+2^-(1/16)]

(1+1/2)(1+1/2^2)(1+1/2^4)(1+1/2^8)(1+1/2^16),(1+1/2)(1+1/2^2)(1+1/2^4)(1+1/2^8)(1+1/2^16),(1+1/2)(1+

计算:1/2-(1/2-1/4)-(1/4-1/8)-…-1/512-1/1024)计算:1/2-(1/2-1/4)-(1/4-1/8)-…-1/512-1/1024)计算:1/2-(1/2-1/4)

1/2-(1/2-1/4)-(1/4-1/8)-.-(1/512-1/1024)1/2-(1/2-1/4)-(1/4-1/8)-.-(1/512-1/1024)1/2-(1/2-1/4)-(1/4-1

2/1-(2/1-4/1)-(4/1-8/1).(8192/1-16384/1)2/1-(2/1-4/1)-(4/1-8/1).(8192/1-16384/1)2/1-(2/1-4/1)-(4/1-8

1/2-(1/2-1/4)-(1/4-1/8)-……-(1/8192-1/16384)1/2-(1/2-1/4)-(1/4-1/8)-……-(1/8192-1/16384)1/2-(1/2-1/4)-

1/2-(1/2-1/4)-(1/4-1/8)-...-(1/8192-1/16284)咋算哪?5555555.1/2-(1/2-1/4)-(1/4-1/8)-...-(1/8192-1/16284)

(2+1)({2}^{2}+1)({2}^{4}+1)({2}^{8}+1).({2}^{64}+1)+1(2+1)({2}^{2}+1)({2}^{4}+1)({2}^{8}+1).({2}^{64