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Cho S= 1-1/2+1/3-1/4+…+1/2011-1/2012+1/2013 và P= 1/1007+1/1008+…+1/2013 Tính (S-P)^2013
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Đáp án:
Ta có :
`S = 1 – 1/2 + 1/3 – 1/4 + …. + 1/2011 – 1/2012 + 1/2013`
`= (1 + 1/3 + …. + 1/2011 + 1/2013) – (1/2 + 1/4 + …. + 1/2012)`
`= (1 + 1/2 + 1/3 + 1/4 + …. + 1/2012 + 1/2013) – 2(1/2 + 1/4 + …. + 1/2012)`
`= (1 + 1/2 + …. + 1/2012 + 1/2013) – (1 + 1/2 + …. + 1/1006)`
`= 1/1007 + 1/1008 + …. + 1/2013`
`=> S – P = (1/1007 + 1/1008 + …. + 1/2013) – (1/1007 + 1/1008 + …. + 1/2013) = 0`
`=> (S – P)^{2013} = 0^{2013} = 0`
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