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

tao có:

2p=2/1.2.3+2/2.3.4+...+2/n.n(+1)n(n+2)

2p=3-1/1.2.3+4-2/1.2.3+...+(n+2)-n/n.(n+1).(n+2)

2p=3/1.2.3-1/1.2.3+4/2.3.4-2/2.3.4+...+(n+2)/n.(n+1).(n+2)-n/n.(n+1).(n+2)

2p=1/1.2-1/2.3+1/2.3-1/3.4+...+1/n.(n+1)-1/(n+1).(n+2)

2p=1/1.2-1/(n+1).(n+2)

2p=(n+!).(n+2)-2/(2n+2).(n+2)

suy ra p=(n+1).(n+2)-2/(2n+2).(2n+4)

2s=3-1/1.2.3+4-2/1.2.3+...+50-48/48.49.50

2s=3/1.2.3-1/1.2.3+4/2.3.4-2/2.3.4+...+50/49.50.48-48/48.50.49

2s=1/1.2-1/2.3+1/2.3-1/3.4+...+1/48.49-1/49.50

2s=1/1.2-1/49.50

'2s=1/2-1/2450

2s=1225/2450-1/2450

2s=1224/2450

s=612/1225

\(P=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{n\left(n+1\right)\left(n+2\right)}\)1

\(P=\frac{1}{2}\left(\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+\frac{2}{3\cdot4\cdot5}+...+\frac{2}{n\left(n+1\right)\left(n+2\right)}\right)\)

\(P=\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+\frac{1}{3\cdot4}-\frac{1}{4\cdot5}+...+\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right)\)

\(P=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right)\)

\(P=\frac{\left(\frac{1}{2}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right)}{2}\)

S cx tinh giong v

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

Đặt A = 1.2.3 + 2.3.4 + 3.4.5 + ... + 28.29.30

4A = 1.2.3.(4-0) + 2.3.4.(5-1) + 3.4.5.(6-2) + ... + 28.29.30.(31-27)

4A = 1.2.3.4 - 0.1.2.3. + 2.3.4.5 - 1.2.3.4 + 3.4.5.6 - 2.3.4.5 + ... + 28.29.30.31 - 27.28.29.30

4A = 28.29.30.31 - 0.1.2.3

4A = 28.29.30.31

\(A=\frac{28.29.30.31}{4}=7.29.30.31=188790\)

Theo cách tính trên ta dễ dàng tính được:

1.2.3 + 2.3.4 + 3.4.5 + ... + (n - 1).n.(n + 1) = \(\frac{\left(n-1\right).n.\left(n+1\right).\left(n+2\right)}{4}\)

549 + X = 1326
X = 1326 - 549
X = 777
X - 636 = 5618
X = 5618 + 636
X = 6254

549 ,1326 ở đâu zậy bạn  !!! :/

4(1.2.3) = 1.2.3.4 - 0.1.2.3

4(2.3.4) = 2.3.4.5 - 1.2.3.4

4(3.4.5) = 3.4.5.6 - 2.3.4.5

....................................

4(n-1)n(n+1) = (n-1)n(n+1)(n+2) - (n-2)(n-1)n(n+1)

=> 4 B = (n-1)n(n+1)(n+2) => B= (n-1)n(n+1)(n+2):4

4(1.2.3)=1.2.3.4 - 0.1.2.3

4(2.3.4)=2.3.4.5 - 1.2.3.4

4(3.4.5)=3.4.5.6 - 2.3.4.5

.........................

......................................

.......................................

thank

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

4B = 1.2.3.4 + 2.3.4.4 + ... + (n-1)n(n+1).4

= 1.2.3.4 - 0.1.2.3 + 2.3.4.5 - 1.2.3.4 + ... + (n-1)n(n+1)(n+2) - [(n-2)(n-1)n(n+1)]

= (n-1)n(n+1)(n+2) - 0.1.2.3

= (n-1)n(n+1)(n+2)

suy ra \(B = {(n-1)n(n+1)(n+2)\over 4}\)

\(S_n=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{n\left(n+1\right)\left(n+2\right)}\)
\(2S_n=\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{\left(n+2\right)-n}{n\left(n+1\right)\left(n+2\right)}\)
\(2S_n=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\)
\(2S_n=\frac{1}{1.2}-\frac{1}{\left(n+1\right)\left(n+2\right)}\)
\(S_n=\frac{1}{4}-\frac{1}{2\left(n+1\right)\left(n+2\right)}\)

\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right)=\frac{n\left(n+3\right)}{4\left(n+1\right)\left(n+2\right)}\)

Cách của bạn Đỗ Ngọc Hải cũng đúng . Mik có cách khác nè : 

\(S_n=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{n\left(n+1\right)\left(n+2\right)}\)

\(\Rightarrow S_n=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right)\)

\(\Rightarrow S_n=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right)\)

\(\Rightarrow S_n=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right)\)

\(\Rightarrow S_n=\frac{1}{4}-\frac{1}{2\left(n+1\right)\left(n+2\right)}\)

~ Ủng hộ nhé