= 1/2*(1/1*2 - 1/2*3 + 1/2*3 - 1/3*4 + ... + 1/8*9 - 1/9*10) = 1/2*(1/1*2 - 1/9*10)=1/2 * 22/45 = 11/45

2A = \(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+...+\frac{1}{8\cdot9}-\frac{1}{9\cdot10}\)

2A = \(\frac{1}{2}-\frac{1}{90}\)

2A = \(\frac{44}{90}\)

A = \(\frac{22}{90}\)

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

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

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

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

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

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

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

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

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

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

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

#\(Toru\)

B = 1.2.3 + 2.3.4 + ...+ ( n - 1 )n( n + 1 )

 \(\Rightarrow\)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)

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

B = 1.2.3 + 2.3.4 +..+ (n-1).n.(n+1) 

4S = 1.2.3.4 + 2.3.4.4 + 3.4.5.4 +..+ (n-1)n(n+1).4 

ghi dọc cho dễ nhìn: 
(k-1)k(k+1).4 = (k-1)k(k+1)[(k+2) - (k-2)] = (k-1)k(k+1)(k+2) - (k-2)(k-1)k(k+1) 
ad cho k chạy từ 2 đến n ta có: 
1.2.3.4 = 1.2.3.4 
2.3.4.4 = 2.3.4.5 - 1.2.3.4 
3.4.5.4 = 3.4.5.6 - 2.3.4.5 
... 
(n-2)(n-1)n.4 = (n-2)(n-1)n(n+1) - (n-3)(n-2)(n-1)n 
(n-1)n(n+1).4 = (n-1)n(n+1)(n+2) - (n-2)(n-1)n(n+1) 
+ + cộng lại vế theo vế + + (chú ý cơ chế rút gọn) 
4S = (n-1)n(n+1)(n+2) 

=> B = (n-1)n(n+1)(n+2)/4 

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

=1.2.3.(4-0)+2.3.4.(5-1)+...+(n-1)n(n+1).(n+2-n+2)

=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)

Vậy B=\(\frac{\text{(n-1)n(n+1)(n+2)}}{4}\)

B = 1.2.3 + 2.3.4 + ... + (n - 1)n(n + 1)

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

4B = 1.2.3.4 + 2.3.4.(5 - 1) + ... + (n - 1)n(n + 1)[(n + 2) - (n - 2)]

4B = 1.2.3.4 - 1.2.3.4 + 2.3.4.5 + ... + (n - 1)n(n + 1)(n + 2) - (n-2)(n-1)n(n+1)

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

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

s= (2/1.2.3 +2/2.3.4+...+2/98.99.100):2=  (1/1.2-1/2.3+1/2.3-1/3.4+...+1/98.99-1/99.100):2=(1/1.2-1/99.100):2=4949/19800=>S=4949/19800

bài này cô dạy mk rùi, nhưng ko mún viết, mỏi tay

\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\right).x=\frac{23}{45}\)

\(\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right).x=\frac{23}{45}\)

\(\frac{1}{2}\left(\frac{1}{2}-\frac{1}{90}\right).x=\frac{23}{45}\)

\(\Rightarrow x=2\)

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

A=1/1.2.3 + 1/2.3.4 + ...  + 1/23.24.25

2A=2/1.2.3 + 2/2.3.4 + ...  + 2/23.24.25

=1/1.2 - 1/2.3 + 1/2.3 -1/3.4 + .... + 1/23.24 - 1/24.25 

=1/1.2 - 1/24.25

Tớ chỉ giải đến đó thôi còn lại các  bạn cứ bấm máy tính là ra

Bài toán trên áp dụng bài toán tổng quát sau:

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

Suy ra 

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

\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{23.24.25}\)

\(=\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{23.24}-\frac{1}{24.25}\right).\frac{1}{2}\)

\(=\left(\frac{1}{1.2}-\frac{1}{24.25}\right).\frac{1}{2}\)

\(=\frac{299}{1200}\)

\(B=1.2.3+2.3.4+...+\left(n-1\right).n.\left(n+1\right)\)

\(4B=1.2.3.4+2.3.4.\left(5-1\right)+...+\left(n-1\right).n.\left(n+1\right)\left[\left(n+2\right)-\left(n-2\right)\right]\)

\(4B=1.2.3.4+2.3.4.5-1.2.3.4+...+\left(n-1\right).n.\left(n+1\right)\left(n+2\right)-\left(n-2\right)\left(n-1\right).n.\left(n+1\right)\)

\(4B=\left(n-1\right).n.\left(n+1\right)\left(n+2\right)\)

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

Tham khảo nhé~

Ta có: \(B=1.2.3+2.3.4+...+\left(n-1\right).n.\left(n+1\right)\)

\(\Leftrightarrow4B=4.\left[1.2.3+2.3.4+...+\left(n-1\right).n.\left(n+1\right)\right]\)

\(\Leftrightarrow4B=1.2.3.4+2.3.4.4+...+\left(n-1\right).n.\left(n+1\right).4\)

\(\Leftrightarrow4B=1.2.3.4+2.3.4\left(5-1\right)+...+\left(n-1\right)n.\left(n+1\right).\left[\left(n+2\right)-\left(n-2\right)\right]\)

\(\Leftrightarrow4B=1.2.3.4+2.3.4.5-1.2.3.4+...+\left(n-1\right).n.\left(n+1\right).\left(n+2\right)-\left(n-2\right).\)\(\left(n-1\right).n.\left(n+1\right)\)

\(\Leftrightarrow4B=\left(n-1\right).n.\left(n+1\right).\left(n+2\right)\)

\(\Leftrightarrow B=\left(n-1\right).n.\left(n+1\right).\left(n+2\right)\div4\)

Vậy \(B=\left(n-1\right).n.\left(n+1\right).\left(n+2\right)\div4\)