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(2013.2014+2014.2015+2015.2016)(1+\(\frac{1}{3}-1-\frac{1}{3}\))
=(2013.2014+2014.2015+2015.2016).0
=0
a, s1 có 2015 hạng tử
=> s1= (2014:2).-1+2015=1007.(-1)+2015=1008
Lời giải:
a,S1=1+(-2)+3+(-4)+...+(-2014)+2015
=(1-2)+(3-4)+...+(2013-2014)+2015
=-1+(-1)+...+(-1)+2015
=-1.1007+2015
=(-1007)+2015
=1008
b,S2=(-2)+4+(-6)+8+...+(-2014)+2016
=(-2+4)+(-6+8)+...+(-2014+2016)
=2+2+...+2
=2.504
=1008
c,S3=1+(-3)+5+(-7)+...+2013+(-2015)
=(1-3)+(5-7)+...+(2013-2015)
=(-2)+(-2)+...+(-2)
=(-2).504
=-1008
d,S4=(-2015)+(-2014)+(-2013)+...+2015+2016
=(-2015+2015)+...+0+2016
=0+...+0+2016
=2016
STUDY WELL !
nè mình gợi ý cho gọi a= 1-1/2-1/2^2-1/2^3-......... ......1/2^2014 1 / 2^2>1 / 2.3 1/2^3>1/3.4 ................ 1/2^2014<1/2014.2015 nen 1-1/2-1/2^2-1/2^3-.........................1/^2014>1-1/1.2-1/2.3-1/3.4-........................1/2014.2015 a<1-[1-1/2015] a<1-2014/2015 a<1/2015
\(\left(\frac{2013}{2011}+\frac{2014}{2012}+\frac{2015}{2013}\right).\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)=0\)
A = 2016^2015 +1 / 2016^2014+1 < 2016^2015 + 1 + 2015 / 2016^2014 + 1 + 2015
= 2016^2015 + 2016 / 2016^2014 + 2016
= 2016(2016^2014 + 1 ) / 2016(2016^2013 +1)
= 2016^2014 + 1 / 2016^2013 + 1 = B
=> A < B
( 2013 x 2014 +2014 x 2015 + 2015 x 2016 ) x ( 1 + 1/3 - 1 - 1/3 )
= ( 2013 x 2014 + 2014 x 2015 + 2015 x 2016 ) x 0
= 0
gọi \(A=\frac{2015^{2015}+1}{2015^{2016}+1};B=\frac{2015^{2014}+1}{2015^{2015}+1}\)
\(\Rightarrow A=\frac{2015^{2015}+1}{2015^{2016}+1}<\frac{2015^{2015}+2014+1}{2015^{2016}+2014+1}=\frac{2015^{2015}+2015}{2015^{2016}+2015}=\frac{2015\left(2015^{2014}+1\right)}{2015\left(2015^{2015}+1\right)}=\frac{2015^{2014}+1}{2015^{2015}+1}=B\)
Đặt \(A=1-\frac{1}{2}-\frac{1}{2^2}-\frac{1}{2^3}-...-\frac{1}{2^{2014}}\)(1)
=>\(\frac{1}{2}.A=\frac{1}{2}-\frac{1}{2^2}-\frac{1}{2^3}-\frac{1}{2^4}-...-\frac{1}{2^{2015}}\)(2)
Trừ (1) cho (2) theo vế ta được: \(A-\frac{1}{2}.A=1-\frac{1}{2}-\frac{1}{2}+\frac{1}{2^{2015}}\)
(chú ý quy tắc bỏ dấu ngoặc)
hay \(\frac{1}{2}.A=\frac{1}{2^{2015}}\)
=>\(A=\frac{1}{2^{2014}}\)
Vì 0 < 22014 < 22015 => \(\frac{1}{2^{2014}}>\frac{1}{2^{2015}}\) => điều phải chứng minh.
\(\left(2013.2014+2014.2015+2015.2016\right).\left(1+\frac{1}{3}-1-\frac{1}{3}\right)\)
\(=\left(2013.2014+2014.2015+2015.2016\right).0\)
= 0