Tính B=1/1008*2014 + 1/1009*2013 + 1/1010*2012 + ...+ 1/2014*1008
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A=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2013}-\frac{1}{2014}\)
Theo đầu bài ta có:
\(S=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2013}-\frac{1}{2014}+\frac{1}{2015}\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}+\frac{1}{2014}+\frac{1}{2015}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2014}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}+\frac{1}{2014}+\frac{1}{2015}\right)-\left(1+\frac{1}{2}+...+\frac{1}{1007}\right)\)
\(=\frac{1}{1008}+\frac{1}{1009}+...+\frac{1}{2015}\)
\(\Rightarrow S=P\)
Vậy ( S - P )2016 = 02016 = 0
$2015=5.13.31$2015=5.13.31
Ta có: $1.2.....1007=1.2...5....13.....31...1007\text{ chia hết cho }5.13.31=2015$1.2.....1007=1.2...5....13.....31...1007 chia hết cho 5.13.31=2015
$1008.1009.....2004=1008....\left(1010\right)....\left(1014\right)...\left(1023\right)....2004$1008.1009.....2004=1008....(1010)....(1014)...(1023)....2004
$=1008....\left(5.202\right)....\left(13.78\right)....\left(31.33\right)...2004\text{ chia hết cho }5.13.33=2015$=1008....(5.202)....(13.78)....(31.33)...2004 chia hết cho 5.13.33=2015
Do đó tổng 2 số trên chia hết cho 2015.
Ta có:
S - P = (1 - 1/2 + 1/3 -1/4+ ...+ 1/1007 - 1/1008 + ...+ 1/2013 - 1/2014 + 1/2015) - (1/1008 + 1/1009 + ...+1/2014 + 1/2015)
=1 - 1/2 + 1/3 - 1/4 + ... + 1007 -2/1008 - ... - 2/2014
= 1 - 1/2 + 1/3 - 1/4 + ...+ 1/1007 - 2/1008 - 2/1010 - ...- 2/2012 - 2/2014
= 1 - 1/2 + 1/3 - 1/4 + ....+ 1007 - 1/504 - 1/505 - ...- 1/1006 - 1/1007
= 1 - 1/2 + 1/3 - 1/4 + ...1/503 - 1/504 + 1/505 + ...+ 1/1005 - 1/1006 + 1/1007 - 1/504 - 1/505 - ...- 1/1006 - 1/1007
= 1 - 1/2 + 1/3 - 1/4 + ...1/503 - 2/504 - 2/506 - ..- 2/1006
= 1 - 1/2 + 1/3 - 1/4 + ...1/503 - 1/252 - 1/253 - ...- 1/503
Lại tiếp tục như trên, Lẻ mất, chẵn còn => S - P = 0 => (S-P)2015=0