Đáp án:
Ta có :
`S = 1- 1/2 + 1/3 - 1/4 +....+ 1/2011 - 1/2012 + 1/2013`
`S = (1+1/3+1/5+...+1/2013)-(1/2+1/4+1/6+...+1/2012)`
`S = (1+1/2+1/3+1/4+1/5+...+1/2013)-2(1/2+1/4+...+1/2012)`
`S = (1+1/2+1/3+1/4+1/5+...+1/2013)-(1+1/2+...+1/1006)`
`S = 1/1007+1/1008+1/1009+...+1/2013 = P`
`->` S - P = (1/1007+1/1008+1/1009+...+1/2013)-(1/1007+1/1008+1/1009+...+1/2013) = 0`
`-> (S-P)^2013 = 0^2013=0`
`Go od luck!`
Có thắc mắc thì hỏi