A=1-1/2+1/2-1/3+1/3-1/4+...+1/99-1/100

A=1-1/100                            A=99/100                                                                                    B= (1/5.6+1/6/7+...+1/101.102).3                         B=(1/5-1/6+1/6-1/7+...+1/101-1/102).3        B=(1/5-1/102).3                                                 B=97/170                                                            

1) Tính

a) Ta có: \(A=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{99\cdot100}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

\(=1-\dfrac{1}{100}=\dfrac{99}{100}\)

\(\dfrac{2}{2\cdot3}+\dfrac{2}{3\cdot4}+...+\dfrac{2}{99\cdot100}\)

\(=\)\(\dfrac{1\cdot2}{2\cdot3}+\dfrac{1\cdot2}{3\cdot4}+...+\dfrac{1\cdot2}{99\cdot100}\)

\(=\)\(2\cdot\left(\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{99\cdot100}\right)\)

\(=\)\(2\cdot\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\right)\)

\(=\)\(2\cdot\left(\dfrac{1}{2}-\dfrac{1}{100}\right)\)

\(=\)\(2\cdot\dfrac{49}{100}\)

\(=\)\(\dfrac{49}{50}\)

= 1/1 - 1/3 + 1/3 - 1/4 + ... + 1/99 - 1/100

= 1/1 - 1/100

= 99/100

\(A=\dfrac{3}{\left(1.2\right)^2}+\dfrac{5}{\left(2.3\right)^2}+...+\dfrac{4033}{\left(2016.2017\right)^2}\)

\(=\dfrac{3}{1.2^2}+\dfrac{5}{2^2.3^2}+...+\dfrac{4033}{2016^2.2017^2}\)

\(=\dfrac{1}{1}-\dfrac{1}{2^2}+\dfrac{1}{2^2}-\dfrac{1}{3^2}+...+\dfrac{1}{2016^2}-\dfrac{1}{2017^2}\)

\(=1-\dfrac{1}{2017^2}< 1\)

\(\Rightarrow A< 1\left(đpcm\right)\)

Vậy...

c) Đặt \(A=1\cdot2+2\cdot3+3\cdot4+...+99\cdot100\)

Ta có: \(A=1\cdot2+2\cdot3+3\cdot4+...+99\cdot100\)

\(\Leftrightarrow3A=3\cdot\left(1\cdot2+2\cdot3+3\cdot4+...+99\cdot100\right)\)

\(\Leftrightarrow3A=1\cdot2\cdot3+2\cdot3\cdot\left(4-1\right)+3\cdot4\cdot\left(5-2\right)+...+99\cdot100\cdot\left(101-98\right)\)

\(\Leftrightarrow3\cdot A=1\cdot2\cdot3-1\cdot2\cdot3+2\cdot3\cdot4-2\cdot3\cdot4+...+98\cdot99\cdot100-98\cdot99\cdot100+99\cdot100\cdot101\)

\(\Leftrightarrow3\cdot A=99\cdot100\cdot101\)

\(\Leftrightarrow A=33\cdot100\cdot101=333300\)

b) Ta có: \(1+2-3-4+...+97+98-99-100\)

\(=\left(1+2-3-4\right)+\left(5+6-7-8\right)+...+\left(97+98-99-100\right)\)

\(=\left(-4\right)+\left(-4\right)+...+\left(-4\right)\)

\(=-4\cdot25=-100\)

S = 1.2 + 2.3 + 3.4 +...+99.100

3S = 1.2.3 + 2.3.(4 - 1) + 3.4(5 - 2) +...+ 99.100(101 - 98)

3S = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 +...+ 99.100.101 - 98.99.100

3S = 99.100.101

3S = 999900

S = 333300

P = 1 + 3 + 5 + 7 +...+ 2015

P = (2015 + 1)1008 : 2 

P = 1016064

T = 1 + 2 - 3 - 4 + 5 + 6 - 7 - 8 +...+ 97 + 98 - 99 - 100

T = (1 + 2 - 3 - 4) + (5 + 6 - 7 - 8) +...+ (97 + 98 - 99 - 100)

T = (-4) + (-4) +...+ (-4)     

T = (-4)25

T = -100

S=999900

P=1016064

T=-100

S= 333300

quên chưa chia cho 3

 Tính tổng :

a ,1+2+3+..........+2015 

SSH của tổng trên là :

   (2015-1):1+1=2015(SH)

Tổng trên là:

  (2015:2)x(2015+1)=2031120

b, 3+5+7+......+2015

SSH của tổng trên là :

     (2015-3):2+1=1007(SH)

Tổng trên là:

     (1007:2)x(2015+3)=1016063

LƯU ý: SSH=số số hạng nha

a 2029106

b508032

c1679780.53381924

tick đúng cho mk nha

                         a)1+2+3+...........+2015 = 2031120

                          b)3+5+7+...........+2015 = 1016063

                          c)1,2+2,3+3,4+...........2015,2016 =20294103,83

Phần A sai nha phải là: 10 + 13 + ... + 79 + 82

đáp án:

SSH: A = ( 82 - 10 ) : 3 + 1 = 25

Tổng: ( 10 + 82 ) . 25 : 2 = 1150

a) A = 10+13 + ...+79 + 81

A = ( 79+10) x 24 : 2 + 81

A = 89 x 24 : 2 +81

A = 1068+ 81

A= 1149

b) chỗ " 1/12.20" phải là 1/12.13 chứ !

 \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{12.13}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{12}-\frac{1}{13}\)

\(=1-\frac{1}{13}\)

\(=\frac{12}{13}\)

c) \(C=\frac{3}{7}\times\frac{4}{13}+\frac{3}{7}+5\frac{4}{7}\)

\(C=\frac{3}{7}\times\left(\frac{4}{13}+1\right)+\frac{39}{7}\)

\(C=\frac{3}{7}\times\frac{17}{13}+\frac{39}{7}\)

\(C=\frac{51}{91}+\frac{39}{7}\)

\(C=\frac{558}{91}\)