giá trị của A = 1 + 2 +2^2 +....+ 2^1007
A) 2^1008-1
B)2^1009 - 1
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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
A=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2013}-\frac{1}{2014}\)
\(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{2015.2016}\)
\(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{2015}-\frac{1}{2016}\)
\(A=\left(1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+...+\frac{1}{2015}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2016}\right)\)
\(A=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2015}+\frac{1}{2016}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2016}\right)\)
\(A=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2015}+\frac{1}{2016}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{1008}\right)\)
\(A=\frac{1}{1009}+\frac{1}{1010}+\frac{1}{1011}+...+\frac{1}{2015}+\frac{1}{2016}\)
\(\Rightarrow B-A=\left(\frac{1}{1008}+\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2016}\right)-\left(\frac{1}{1009}+\frac{1}{1010}+\frac{1}{1011}+...+\frac{1}{2016}\right)\)
\(\Rightarrow B-A=\frac{1}{1008}\)