4A = 4.[1.2.3 + 2.3.4 + 3.4.5 + … + (n – 1).n.(n + 1)]

4A = 1.2.3.4 + 2.3.4.4 + 3.4.5.4 + … + (n – 1).n.(n + 1).4

4A = 1.2.3.4 + 2.3.4.(5 – 1) + 3.4.5.(6 – 2) + … + (n – 1).n.(n + 1).[(n + 2) – (n – 2)]

4A = 1.2.3.4 + 2.3.4.5 – 1.2.3.4 + 3.4.5.6 – 2.3.4.5 + … + (n – 1).n(n + 1).(n + 2) – (n – 2).(n – 1).n.(n + 1)

4A = (n – 1).n(n + 1).(n + 2)

A = (n – 1).n(n + 1).(n + 2) : 4.

cau a thi sao ha ban ? 

ok thanks ban nhe

Câu a)
\(A=2^{100}-2^{99}+2^{98}-2^{97}+...+2^2-2\)
\(=\left(2^{100}+2^{99}+2^{98}+2^{97}+...+2^2+2\right)-2\left(2^{99}+2^{97}+2^{95}+...+2^3+2\right)\)
\(=\left(2^{100}+2^{99}+2^{98}+2^{97}+...+2^2+2\right)-\left(2^{100}+2^{98}+2^{96}+...+2^4+2^2\right)\)
\(=2^{99}+2^{97}+2^{95}+...+2^3+2\)
\(=\frac{2^2\cdot\left(2^{99}+2^{97}+2^{95}+...+2^3+2\right)-\left(2^{99}+2^{97}+2^{95}+...+2^3+2\right)}{3}\)
\(=\frac{\left(2^{101}+2^{99}+2^{97}+...+2^5+2^3\right)-\left(2^{99}+2^{97}+2^{95}+...+2^3+2\right)}{3}\)
\(=\frac{2^{101}-2}{3}\)

\(2B=\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{2015.2016.2017}\)

\(2B=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{2.4}+...+\frac{1}{2015.2016}-\frac{1}{2016.2017}\)

\(2B=\frac{1}{1.2}-\frac{1}{2016.2017}\)

\(B=\frac{\frac{1}{1.2}-\frac{1}{2016.1017}}{2}\)

B=1.2.3+2.3.4+3.4.5+...+n(n+1)(n+2)

  ={1.2.3.(4-0)+2.3.4(5-1)+3.4.5.(6-2)+...+n(n+1)(n+2)[(n+3)-(n-1)]} : 4

  = [1.2.3.4+2.3.4.5+3.4.5.6+...+n(n+1)(n+2)(n+3) - 1.2.3.4 - 2.3.4.5 - 3.4.5.6 - ... - n(n+1)(n+2)(n-1)] : 4

  =\(\frac{\text{ n(n+1)(n+2)(n+3) }}{4}\)

B = \(\frac{n\left(n+1\right)\left(n+2\right)\left(n+3\right)}{4}\)

Ta có : A = 1.2.3 + 2.3.4 + 4.5.6 + ..... + 98.99.100

=> 6A = 1.2.3.4 - 1.2.3.4 + 2.3.4.5 - 2.3.4.5 + ...... + 98.99.100.101

=> 6A = 98.99.100.101 

=> A = \(\frac{98.99.100.101}{6}=16331700\)

có 2017² đồng dư 1 mod (3)
   => (2017²)⁵⁰ đồng dư 1 mod (3)
=> (2017²)⁵⁰-1 đồng dư 1-1 = 0 mod (3)
=> dpcm

A = \(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2007}}+\frac{1}{3^{2008}}\)

3A= \(1+\frac{1}{3}+...+\frac{1}{3^{2006}}+\frac{1}{3^{2007}}\)

3A-A= \(1-\frac{1}{3^{2008}}\)

B = \(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{n-1}}+\frac{1}{3^n}\)

3B = \(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{n-2}}+\frac{1}{3^{n-1}}\)

3B - B = \(1-\frac{1}{3^n}\)

Ta có :

\(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2007}}+\frac{1}{3^{2008}}\)

\(\Leftrightarrow\)\(3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2006}}+\frac{1}{3^{2007}}\)

\(\Leftrightarrow\)\(3A-A=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2006}}+\frac{1}{3^{2007}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2007}}+\frac{1}{3^{2008}}\right)\)

\(\Leftrightarrow\)\(2A=1-\frac{1}{3^{2008}}\)

\(\Leftrightarrow\)\(2A=\frac{3^{2008}-1}{3^{2008}}\)

\(\Leftrightarrow\)\(A=\frac{3^{2008}-1}{3^{2008}}:2\)

\(\Leftrightarrow\)\(A=\frac{3^{2008}-1}{2.3^{2008}}\)

Vậy \(A=\frac{3^{2008}-1}{2.3^{2008}}\)

https://olm.vn/hoi-dap/tim-kiem?q=t%C3%ADnh+t%E1%BB%95ng+sau+:S+=+1.2.3+2.3.4+3.4.5+...+n.(n+1).(n+2)+&id=601088

\(\dfrac{1}{1\cdot2\cdot3}+\dfrac{1}{2\cdot3\cdot4}+\cdot\cdot\cdot+\dfrac{1}{\left(n-2\right)\cdot\left(n-1\right)\cdot n}\)

\(=\dfrac{1}{2}\left(\dfrac{2}{1\cdot2\cdot3}+\dfrac{2}{2\cdot3\cdot4}+\cdot\cdot\cdot+\dfrac{2}{\left(n-2\right)\cdot\left(n-1\right)\cdot n}\right)\)

\(=\dfrac{1}{2}\left[\dfrac{1}{1\cdot2}-\dfrac{1}{2\cdot3}+\dfrac{1}{2\cdot3}-\dfrac{1}{3\cdot4}+\cdot\cdot\cdot+\dfrac{1}{\left(n-2\right)\left(n-1\right)}-\dfrac{1}{\left(n-1\right)n}\right]\)

\(=\dfrac{1}{2}\left[\dfrac{1}{1\cdot2}-\dfrac{1}{\left(n-1\right)n}\right]\)

\(=\dfrac{1}{2}\cdot\left[\dfrac{n\left(n-1\right)}{2n\left(n-1\right)}-\dfrac{2}{2n\left(n-1\right)}\right]\)

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

\(=\dfrac{n^2-n-2}{4n\left(n-1\right)}\)

#\(Toru\)

https://hoc247.net/hoi-dap/toan-6/tinh-tong-s-1-1-2-3-1-2-3-4-1-n-n-1-n-2--faq240420.html

`->` Mình tham khảo ở đây để làm nếu sai thì cho mik xl ạ.

Đặt \(A=\dfrac{1}{1\cdot2\cdot3}+\dfrac{1}{2\cdot3\cdot4}+....+\dfrac{1}{\left(n-2\right)\cdot\left(n-1\right)\cdot n}\)

\(2A=\dfrac{2}{1\cdot2\cdot3}+\dfrac{2}{2\cdot3\cdot4}+...+\dfrac{2}{\left(n-2\right)\cdot\left(n-1\right)\cdot n}\\ 2A=\dfrac{1}{1\cdot2}-\dfrac{1}{2\cdot3}+\dfrac{1}{2\cdot3}-\dfrac{1}{3\cdot4}+....+\dfrac{1}{\left(n-2\right)\cdot\left(n-1\right)}-\dfrac{1}{\left(n-1\right)\cdot n}\)

\(2A=\dfrac{1}{1\cdot2}-\dfrac{1}{\left(n-1\right)\cdot\left(n-2\right)}\)

\(A=\dfrac{1}{4}-\dfrac{1}{\left(n-1\right)\cdot\left(n-2\right)\cdot2}\)

Mình làm mẫu 1 bài nha !

Có : 12A = 1.5.12+5.9.12+....+101.105.12

= 1.5.12+5.9.(13-1)+.....+101.105.(109-97)

= 1.5.12+5.9.13-1.5.9+.....+101.105.109-97.101.105

= 1.5.12-1.5.9+101.105.109

= 1155960

=> A = 1155960 : 12 = 96330

Tk mk nha

Có : 4D = 1.2.3.4+2.3.4.4+....+98.99.100.4

= 1.2.3.4+2.3.4.(5-1)+.....+98.99.100.(101-97)

= 1.2.3.4+2.3.4.5-1.2.3.4+......+98.99.100.101-97.98.99.100

= 98.99.100.101

=> D = 98.99.100.101/4 = 24497550

C = 1.2.3+ 2.3.4 + 3.4.5 +...+n(n+1) ( n+2)

\(\Rightarrow4C=1.2.3\left(4-0\right)+2.3.4.\left(5-1\right)+...+n\left(n+1\right)\left(n+2\right)\left[\left(n+3\right)-\left(n-1\right)\right]\)

            \(=1.2.3.4-0.1.2.3+2.3.4.5-...+n\left(n+1\right)\left(n+2\right)\left(n+3\right)-\left(n-1\right)n\left(n+1\right)\left(n+2\right)\) \(=n\left(n+1\right)\left(n+2\right)\left(n+3\right)-0.1.2.3\)

 \(=n\left(n+1\right)\left(n+2\right)\left(n+3\right)\)

\(\Rightarrow C=\frac{n\left(n+1\right)\left(n+2\right)\left(n+3\right)}{4}\)

Thanks