2 câu trả lời
`S_1 = 1 +1/2 + 1/2^2 + 1/2^3 + .... + 1/2^2006`
`2 . S_1 = 2 + 1 + 1/2 + 1/2^2 + .... + 1/2^2005`
`2 . S_1 - S = ( 2 + 1 + 1/2 + 1/2^2 + .... + 1/2^2005 ) - ( 1 +1/2 + 1/2^2 + 1/2^3 + .... + 1/2^2006 )`
`S_1 = 2 - 1/2^2006`
Vậy `, S_1 = 2 - 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)`