\(M=\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}.....\frac{10^2}{10.11}\)

\(M=\frac{1.1}{1.2}.\frac{2.2}{2.3}.\frac{3.3}{3.4}......\frac{10.10}{10.11}\)

\(M=\frac{1.2.3.....10}{1.2.3....10}.\frac{1.2.3.....10}{2.3.4.....11}\)

\(M=1.\frac{1}{11}\)

\(M=\frac{1}{11}\)

Cảm ơn bạn nha

A=1.2+2.3+3.4+4.5+5.6+...+2016.2017 

=> 3A = 1.2.3+2.3.3+3.4.3+4.5.3+5.6.3+.......+2016.2017.3

=> 3A = 1.2.3 + 2.3.(4-1) + 3.4.(5-2) + 4.5.(6-3) + .......+ 2016.2017.(2018-2015)

=> 3A = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 +..........+ 2016.2017.2018 - 2015.2016.2017

=> 3A = 2016.2017.2018

=> A = 2016.2017.2018 : 3 

Mày.ngu qua

\(A=1.2+2.3+3.4+...+2016.2017\)

\(3A=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+2016.2017.\left(2018-2015\right)\)

\(3A=1.2.3+2.3.4-1.2.3+3.4.5-2.5.4+...+2016.2017.2018-2015.2016.2017\)

\(A=\frac{2016.2017.2018}{3}==2735245632\)

a=1.2 + 2.3 +3.4+ ...+ 2010.2011\

3a=1.2.3+2.3.3+3.4.3+......+2010.2011.3

3a=1.2.3+2.3.(4-1)+3.4.(5-1)+............+2010.2011.(2012-2009)

3a=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+....+2010.2011.2012-2009.2010.2011

3a=2010.2011.2012

a=2010.2011.2012:3

a=?

A=2010.2011.2012:3

a) Đặt A = 1/2 + 1/4 + 1/8 + 1/16 + 1/32

A = 1/2 + 1/2² + 1/2³ + 1/2⁴ + 1/2⁵

2A = 2(1/2 + 2² + 1/2³ + 1/2⁴ + 1/2⁵)

2A = 1 + 1/2 + 1/2² + 1/2³ + 1/2⁴

2A - A = (1 + 1/2 + 1/2²  + 1/2³ + 1/2⁴) - (1/2  + 1/2²  + 1/2³ + 1/2⁴ + 1/2⁵)

A = 1 - 1/2⁵

A = 31/32

b) 2/1.2 + 2/2.3 + 2/3.4 + ... + 2/18 . 19 + 2/19.20

= 2(1/1.2  + 1/2.3  + 1/3.4 + ... + 1/18.19 + 1/19.20)

= 2.(1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/18 - 1/19 + 1/19 - 1/20)

= 2. (1 - 1/20)

= 2.19/20

= 19/10

bạn thiếu ý C 

Sửa đề câu c) 

2 - 4 + 6 - 8 + 10 - 12 + ...... + 98 - 100 + 102 

= 2 + ( 6 - 4 ) + ( 10 - 8 ) + ..... + ( 98 - 96 ) + ( 102 - 100 )

= 2 + ( 2 + 2 + 2 + 2 + .... + 2 )

= 2 + 50

= 52

\(A=1\cdot2+2\cdot3+...+151\cdot152\)

\(=1\left(1+1\right)+2\left(1+2\right)+...+151\left(1+151\right)\)

\(=\left(1+2+3+...+151\right)+\left(1^2+2^2+...+151^2\right)\)

\(=\dfrac{151\left(151+1\right)}{2}+\dfrac{151\left(151+1\right)\left(2\cdot151+1\right)}{6}\)

\(=151\cdot76+\dfrac{151\cdot152\cdot303}{6}\)

\(=151\cdot76+151\cdot7676=1170552\)

\(C=2\cdot4+4\cdot6+...+2024\cdot2026\)

\(=2\cdot2\left(1\cdot2+2\cdot3+...+1012\cdot1013\right)\)

\(=4\left[1\left(1+1\right)+2\left(1+2\right)+...+1012\left(1+1012\right)\right]\)

\(=4\left[\left(1+2+...+1012\right)+\left(1^2+2^2+...+1012^2\right)\right]\)

\(=4\left[1012\cdot\dfrac{1013}{2}+\dfrac{1012\left(1012+1\right)\left(2\cdot1012+1\right)}{6}\right]\)

\(=4\left[506\cdot1013+345990150\right]\)

\(=1386010912\)

\(M=1^2+2^2+...+2024^2\)

\(=\dfrac{2024\left(2024+1\right)\cdot\left(2\cdot2024+1\right)}{6}\)

\(=2024\cdot2025\cdot\dfrac{4049}{6}\)

=2765871900

\(N=1^3+2^3+...+100^3\)

\(=\left(1+2+3+...+100\right)^2\)

\(=\left[\dfrac{100\left(100+1\right)}{2}\right]^2\)

\(=\left[50\cdot101\right]^2=5050^2\)

\(Q=1^3+2^3+...+2024^3\)

\(=\left(1+2+3+...+2024\right)^2\)

\(=\left[\dfrac{2024\left(2024+1\right)}{2}\right]^2\)

\(=\left[1012\left(2024+1\right)\right]^2\)

\(=2049300^2\)

\(A=\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2016.2017}\right):2\)

\(=\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2016}-\frac{1}{2017}\right):2\)

\(=\left(1-\frac{1}{2017}\right):2\)\(< \)\(\frac{1}{2}\)   (Do 1 - 1/2017 < 1)