Tính : S1= 1+1/2+1/2^2=1/2^3+...+1/2^2006

Đáp án:

`S_{1} = ( 2^2007 - 1 )/(2^2006)`

Giải thích các bước giải:

`S_{1} = 1 + 1/2 + 1/(2^2 ) + ... + 1/(2^2006)`

`=> 2S_{1} = 2 . ( 1 + 1/2 + 1/(2^2 ) + ... + 1/(2^2006) )`

`=> 2S_{1} = 2 + 1 + 1/2 + 1/(2^2 ) + ... + 1/(2^2005)`

`=> 2S_{1} - S_{1} = ( 2 + 1 + 1/2 + ... + 1/(2^2005)) - ( 1 + 1/2 + 1/(2^2 ) + ... + 1/(2^2006))`

`=> S_{1} = 2 + 1 + 1/2 + ... + 1/(2^2005) - 1 - 1/2 - 1/(2^2 ) - ... - 1/(2^2006)`

`=> S_{1} = 2 + ( 1 - 1 ) + ( 1/2 - 1/2 ) + ... + 1/(2^2005) - 1/(2^2005) - 1/(2^2006)`

`=> S_{1} = 2 - 1/(2^2006)`

`=> S_{1} = ( 2 . 2^2006 - 1 )/(2^2006)`

`=> S_{1} = ( 2^2007 - 1 )/(2^2006)`

Vậy `S_{1} = ( 2^2007 - 1 )/(2^2006)`