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![](https://rs.olm.vn/images/avt/0.png?1311)
a) A = 1 - 2 + 3 - 4 + .....+ 2013 - 2014 + 2015
= (1 - 2) + (3 - 4) + .....+ (2013 - 2014) + 2015
=(-1) + (-1) + ...+ (-1) + 2015
1007 số -1
=(-1) . 1007 +2015
= -1007 + 2015
=1008
b) 1 . 2 + 2 . 3 + 3 . 4 +....99 . 100
đặt S= 1 . 2 + 2 . 3 + 3 . 4 +....99 . 100
=> 3S = 1.2.3 +2.3.(4-1) +...+99.100.(101-98)
= 1.2.3 +2.3.4-1.2.3+...+99.100.101-98.99.100
= 99.100.101
= 999900
=> S= 333300
Vậy 1 . 2 + 2 . 3 + 3 . 4 +....99 . 100 = 333300
a,A=(1-2)+(3-4)....(2013-2014)+2015
A= -1 + -1.....-1+2015
A= (2015-1):1+1
A=2015
A=(2015 x -1) x -1
A=2015
A=2015 + 2015
A=4030
b, 1/1.2 +1/2.3 ...1/99.100
1/1-1/2+1/2-1/3 ....1/99-1/100
1/1-1/100
99/100
![](https://rs.olm.vn/images/avt/0.png?1311)
C = 1+1/2(1+2)+1/3(1+2+3)+........+1/2015(1+2+3+4+...+2015)
C = 1 + \(\frac{1}{2}\cdot\frac{2.3}{2}\)+ \(\frac{1}{3}\cdot\frac{3.4}{2}\)+ ... + \(\frac{1}{2015}\cdot\frac{2015.2016}{2}\)
C = \(\frac{2}{2}\) + \(\frac{3}{2}+\frac{4}{2}+...+\frac{2016}{2}\)
C = \(\frac{2+3+4+...+2016}{2}\)
Đặt D = 2 + 3 + 4 + ... + 2016
Số số hạng của D là : (2016 - 2) : 1 + 1 = 2015
Tổng D là : (2 + 2016) . 2015 : 2 = 2033135
Thay D vào biểu thức C ta được : \(\frac{2033135}{2}\)
Vậy C = ... .
![](https://rs.olm.vn/images/avt/0.png?1311)
a)\(2S=2\left(1+\frac{1}{2}+...+\frac{1}{2^{100}}\right)\)
\(2S=2+1+...+\frac{1}{2^{99}}\)
\(2S-S=\left(2+1+...+\frac{1}{2^{99}}\right)-\left(1+\frac{1}{2}+...+\frac{1}{2^{100}}\right)\)
\(S=2-\frac{1}{2^{100}}\)
phần b tương tự
a. S=1+1/2+1/2^2+1/2^3+...+1/2^100
2S=2+1+1/2+1/2^2+...+1/2^99
2S-S=(2+1+1/2+1/2^2+...+1/2^99)-(1+1/2+1/2^2+1/2^3+...+1/2^100)
S=2-1/2^100
S=2^101-1/2^100
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{B}{A}=\frac{\frac{2016}{1}+\frac{2015}{2}+...+\frac{2}{2015}+\frac{1}{2016}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+..+\frac{1}{2016}+\frac{1}{2017}}\)
\(\frac{B}{A}=\frac{\left(\frac{2016}{1}+1\right)+\left(\frac{2015}{2}+1\right)+...+\left(\frac{1}{2016}+1\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}+\frac{1}{2017}}\)
\(\frac{B}{A}=\frac{\frac{2017}{1}+\frac{2017}{2}+...+\frac{2017}{2016}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}+\frac{1}{2017}}\)
\(\frac{B}{A}=\frac{2017\cdot\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}+\frac{1}{2017}}=2017\div\frac{1}{2017}=4068289\)