Tính tổng: S= (- 1) + (- 1) ^ 2 + (- 1) ^ 3 +...+(-1)^ 2014 +(-1)^ 2015
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
R.2=2+2^2+2^3+...+2^2015
R=(2+2^2+2^3+...+2^2015-1)-(1+2^2+2^3+...+2^2014)
R=(2^2015)-1
S=(2^2015)-1 / 1-(2^2015)
S=-1
bieu thuc do goi la R nhe
![](https://rs.olm.vn/images/avt/0.png?1311)
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 !
![](https://rs.olm.vn/images/avt/0.png?1311)
Đảo ngược số từ dưới lên ta có.
S=2015-2014+2013-2012+......+7-6+5-4+3-2+1
S= 1 + 1 +.......+ 1 + 1 + 1 +1 (1008 số 1)
S= 1 x 1008 = 1008
Vậy S = 1008
![](https://rs.olm.vn/images/avt/0.png?1311)
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 \(P=\frac{1}{1008}+\frac{1}{1009}+...+\frac{1}{2015}\)
\(\Leftrightarrow S-P=0\) \(\Rightarrow\left(S-P\right)^{2016}=0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
S = 2020 + 2019 - 2018 - 2017 + 2016 + 2015 - 2014 - 2013 + ... + 4 + 3 - 2 - 1
= ( 2020 + 2019 - 2018 - 2017 ) + ( 2016 + 2015 - 2014 - 2013 ) + ... + ( 4 + 3 - 2 - 1 ) (có tất cả 2020 : 4 = 505 nhóm)
= 4 + 4 + ... + 4
= 4. 505 = 2020
Vậy S = 2020.
![](https://rs.olm.vn/images/avt/0.png?1311)
a
so so hang
(100-1):1+1=100(so hang)
tong bang
(100+1)x100:2=5050
\(S=-2\)
\(\cdot DuyNam\)
\(S=\left(-1\right)+\left(-1\right)^2+\left(-1\right)^3+...+\left(-1\right)^{2014}+\left(-1\right)^{2015}\)
\(S=\left(-1\right)+1+\left(-1\right)+...+1+\left(-1\right)\) (2015 thừa số)
`-> S= (-1)`