a) \(A=\frac{1}{8}+\frac{1}{24}+\frac{1}{48}+...+\frac{1}{10200}\)

\(A=\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{100.102}\)

\(2A=\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{100.102}\)

\(2A=\left(\frac{1}{2}-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{6}\right)+\left(\frac{1}{6}-\frac{1}{8}\right)+...+\left(\frac{1}{100}-\frac{1}{102}\right)\)

\(2A=\frac{1}{2}-\frac{1}{102}\)

\(2A=\frac{25}{51}\)

\(A=\frac{25}{51}:2\)

\(A=\frac{25}{102}\)

Vậy \(\frac{1}{8}+\frac{1}{24}+\frac{1}{48}+...+\frac{1}{10200}=\frac{25}{102}\)

b) \(B=\frac{3}{1.2}+\frac{3}{2.3}+\frac{3}{3.4}+...+\frac{3}{2015.2016}\)

\(B=3.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2015.2016}\right)\)

\(B=3.\left[\left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+...+\left(\frac{1}{2015}-\frac{1}{2016}\right)\right]\)

\(B=3.\left(\frac{1}{1}-\frac{1}{2016}\right)\)

\(B=3.\frac{2015}{2016}\)

\(B=\frac{2015}{672}\)

Vậy \(\frac{3}{1.2}+\frac{3}{2.3}+\frac{3}{3.4}+...+\frac{3}{2015.2016}=\frac{2015}{672}\)

bó tay kkk

B = 6 + 6^3 + 6^5 + ... + 6^2015

=> 6^2.B = 6^2(6 + 6^3 + 6^5 + ... + 6^2015

=> 36B = 6^2.6 + 6^3.6 + 6^5.6 + ... + 6^2015 .6 

=> 36B = 6^3 + 6^4 + 6^6 + ... + 6^2016

Lấy 36B trừ đi B, ta có:

     35B = 6^2016 - 6 

=> B = (6^2016 - 6)/35

A= 1.2+2.3+3.4+...+2015.2016

3A=1.2.3+2.3.3+3.4.3+...+2015.2016.3

3A=1.2.3+2.3.(4-1)+3.4.(5-2)+...+2015.2016.(2017-2014)

3A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+2015.2016.2017-2014.2015.2016

3A=2015.2016.2017

3A=8193538080

A=8193538080:3

A=2731179360

3A = 1.2.3 + 2.3.3 + 3.4.3 + ..... + 2015.2016.3

=> 3A = 1.2.3 + 2.3.( 4 -1 ) + 3.4.( 5 - 2 ) + .... + 2015.2016.( 2017 - 2014 )

=> 3A = 1.2.3 + 2.3.4 - 1.2.3 + .... + 2015.2016.2017 - 2014.2015.2016

=> 3A = 2015.2016.2017

=> A = \(\frac{2015.2016.2017}{3}\)

Đặt \(A=1.2+2.3+3.4+...+2015.2016\)

\(\Rightarrow3A=1.2.3+2.3.3+3.4.3+...+2015.2016.3\)

\(\Rightarrow3A=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+2015.2016.\left(2017-2014\right)\)

\(\Rightarrow3A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+2015.2016.2017-2014.2015.2016\)

\(\Rightarrow3A=2015.2016.2017\)

\(\Rightarrow A=2015.2016.2017:3\)

\(\Rightarrow A=2015.672.2017\)

Vậy \(A=2015.672.2017\)

1 . 2 + 2 . 3 + 3 . 4 + ... + 2015 . 2016

3M = 1 . 2 . 3 + 2 . 3 . 3 + 3 . 4 . 3 + ... + 2015 . 2016 . 3 

3M = 1 . 2 ( 3 - 0 ) + 2 . 3 ( 4 - 1 ) + 3 . 4 ( 5 - 2 ) + ... + 2015 . 2016 ( 2017 - 2014 )

3M = ( 1 . 2 . 3 + 2 . 3 . 4 + 3 . 4. 5 + ... + 2015 . 2016 . 2017 ) - ( 0 . 1 . 2 + 1 . 2 . 3 + 2 . 3 . 4 + ... + 2014 . 2015 . 2016 )

3M = 2015 . 2016 . 2017

M = \(\frac{2015.2016.2017}{3}\)

M = 2731179360

A=1.2+2.3+3.4+...+2015.2016

=> 3A=1.2.3+2.3.3+3.4.3+...+2015.2016.3

=> 3A=1.2.3+2.3.(4-1)+3.4.(5-2)+...+2015.2016.(2017-2014)

=>3A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+ 2015.2016.2017-2014.2015.2016

=> 3A=2015.2016.2017

=> A=\(\frac{2015.2016.2017}{3}=2731179360\)

.

.

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

A= 1.2+2.3+3.4+...+2015.2016

3A=1.2.3+2.3.3+3.4.3+...+2015.2016.3

    =1.2.3+2.3.(4-1)+3.4.(5-2)+...+2015.2016.(2017-2014)

    =1.2.3-1.2.3+2.3.4-2.3.4+3.4.5+...-2014.2015.2016+2015.2016.2017

    =2015.2016.2017

A=2015.2016.2017:3=2731179360

\(S=\dfrac{3}{1.2}+\dfrac{3}{2.3}+...+\dfrac{3}{2015.2016}\)

\(=3\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{2015.2016}\right)\)

\(=3\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2015}-\dfrac{1}{2016}\right)\)

\(=3\left(1-\dfrac{1}{2016}\right)\)

\(=3.\dfrac{2015}{2016}=\dfrac{6045}{2016}\)

Vậy \(S=\dfrac{6045}{2016}\)

\(S=3\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2015.2016}\right)\)

\(\Rightarrow S=3\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2015}-\dfrac{1}{2016}\right)\)

\(\Rightarrow S=3\left(1-\dfrac{1}{2016}\right)=3.\dfrac{2015}{2016}=\dfrac{6045}{2016}\)

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: \(1+2-3-4+...+97+98-99-100\)

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

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

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