(1)当n=1时 左边= 右边==1/2,等式成立
(2)假设当n=k时,等式成立即1/2+1/4+1/8+...+1/(2^k) = 1-1/(2^k)
则当n=k+1时1/2+1/4+1/8+...+1/(2^k)+ 1/[2^(k+1)]= 1-1/(2^k)+1/[2^(k+1)] =1-1/[2^(k+1)]
故命题成立
(1)当n=1时 左边= 右边==1/2,等式成立
(2)假设当n=k时,等式成立即1/2+1/4+1/8+...+1/(2^k) = 1-1/(2^k)
则当n=k+1时1/2+1/4+1/8+...+1/(2^k)+ 1/[2^(k+1)]= 1-1/(2^k)+1/[2^(k+1)] =1-1/[2^(k+1)]
故命题成立