Tinh tong S=1/1×2+1/2×3+1/3×4+...+1/2014×2015+1/2015×2016
Tinh tong S=1/1×2+1/2×3+1/3×4+...+1/2014×2015+1/2015×2016
S = 1/1x2 + 1/2x3 + 1/3x4 + ... + 1/2014x2015 + 1/2015x2016
S = 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/2014 - 1/2015 + 1/2015 - 1/2016
S = 1 - 1/2016
S = 2015
\(S=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-......+\frac{1}{2015}-\frac{1}{2016}\)
\(S=\frac{1}{1}-\frac{1}{2016}=\frac{2015}{2016}\)
tinh tong :S=1-2+3-4+5-6+...+2015-2016
S= 1-2 + 3-4 + 5-6+... ..+2015-2016(có 2016 số)
=(1-2) + (3-4) + (5-6) +...+(2015-2016) (có 2016:2=1008 nhóm có 2 số)
=-1 +(-1) +(-1) +...+(-1)( có 1008 số(-1))
=-1.1008
=-1008
vậy S=-1008
tìm s
Tính S = 1/2(1+2) + 1/3(1+2+3)+...+ 1/2015(1+2+...+2014+2015) + 1/2016(1+2+...+2015+2016
Tính S = 1/2(1+2) + 1/3(1+2+3)+...+ 1/2015(1+2+...+2014+2015) + 1/2016(1+2+...+2015+2016)
Tinh:
S=2015 + 2015/1+2 +2015/1+2+3 + 2015/1+2+3+4 +... + 2015/1+2+3+...+2016
Tinh:
S=2015 + 2015/1+2 +2015/1+2+3 + 2015/1+2+3+4 +... + 2015/1+2+3+...+2016
tính các tổng sau
a, S1=1+(-2)+3+(-4)+..........+(-2014)+2015
b,S2=(-2)+4+(-6)+8+...............+(-2014)+2016
c,S3=1+(-3)+5+(-7)+................+2013+(-2015)
d,S4=(-2015)+(-2014)+(-2013)+......+2015+2016
làm đầy đủ chắc chắn cho mk nhé !
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 !
Cô ơi dấu hiệu chia hết cho 5 em mở không được
cho S = 1-1/2+1/3-1/4+...-1/2014+1/2015
P = 1/1008+1/1009+...+1/2014+1/2015
tính (S-P)^2016
Ta có:
\(S=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...-\frac{1}{2014}+\frac{1}{2015}=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{2015}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2014}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}+\frac{1}{2015}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2014}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}\right)-\left(1+\frac{1}{2}+...+\frac{1}{1007}\right)=\frac{1}{1008}+\frac{1}{1009}+....+\frac{1}{2015}\)
Mà \(P=\frac{1}{1008}+\frac{1}{1009}+...+\frac{1}{2015}\)
\(\Leftrightarrow S-P=0\) \(\Rightarrow\left(S-P\right)^{2016}=0\)
tinh A/B
A=1/2+1/3+...+1/2016
B=1/2015+2/2014+...+2014/2+2015/1