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2, 2^ -1 + 2^ -2 + 2^ -3 + .......+ 2^ -18 + 2^ -19
= 2^(-19) *( 2^18 + 2^17+...+2+1)
= 2^(-19) * ( 2^19 -1)/(2-1)
= 1-2^(-19)

1. (X-A)(X-B)(x-C)........(X-X)(X-Y)(X-Z)
= 0 since X - X = 0
2. Let S = 2^-1+2^-2+2^-3+............+2^-18+2^-19 ............(1)
2S = 1+2^-1+2^-2+2^-3+............+2^-18 ......................(2)
(2) - (1)
2S - S = 1 -2^-19
S = 1 - 2^-19

26x-(a+b+c+d+e+f+g+h+i+g+k+l+m+n+o+p+q+r+s+t+u+v+w+x+y+z)

1) ( X-A )( X-B )( X -C )( X-D )..........( X-Z )

( X-A )( X-B )( X -C )( X-D )..........( X-Z )
= ( X-A )( X-B )( X -C )( X-D )..........(X-X)(X-Y)( X-Z )
=( X-A )( X-B )( X -C )( X-D )..........(0)(X-Y)( X-Z )
=0

2. 2^ -1 + 2^ -2 + 2^ -3 + .......+ 2^ -18 + 2^ -19
2的 負1次方加 2的 負2次方加 2的 負3次方......加2的 負18次方 加2的 負19次方

Same as above, but you can also do it as a geometric series:
Let R=1/2, then
2^ -1 + 2^ -2 + 2^ -3 + .......+ 2^ -18 + 2^ -19
=R(1+R+R^2+...+R^18)
=R(1-R^19)/(1-R)
=1-R^19
=1-(1/2)^19

2007-11-28 10:16:13 補充：
假設 a=1/2, 并且 s=1+a+a^2+a^3+....+a^n (=總和)a*s = a+a^2+a^3+a^4+...+a^(n+1)s-a*s = 1-a^(n+1)s(1-a)=1-a^(n+1)s=(1-a^(n+1))/(1-a)可以睇下 http://baike.baidu.com/view/382.html