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

  = 1+(n+1)

  =n+2

B=1

C=3459652

D=5482

E=54235452

F=547754

G=1456154

HUYỀN MUỐN NÊU RA CÁCH TÍNH CHỨ KHÔNG PHẢI KẾT QUẢ

đúng đi đã

a)    \(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{98.99.100}\)

\(=\frac{1}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{98.99.100}\right)\)

\(=\frac{1}{2}\left(\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+...+\frac{100-98}{98.99.100}\right)\)

\(=\frac{1}{2}\left(\frac{3}{1.2.3}-\frac{1}{1.2.3}+\frac{4}{2.3.4}-\frac{2}{2.3.4}+...+\frac{100}{98.99.100}-\frac{98}{98.99.100}\right)\)

\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)\)

\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{9900}\right)\)

\(=\frac{1}{4}-\frac{1}{19800}=\frac{4949}{19800}\)

b) \(\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+...+\frac{1}{27.28.29.30}\)

\(=\frac{1}{3}\left(\frac{3}{1.2.3.4}+\frac{3}{2.3.4.5}+...+\frac{3}{27.28.29.30}\right)\)

\(=\frac{1}{3}(\frac{4-1}{1.2.3.4}+\frac{5-2}{2.3.4.5}+...+\frac{30-27}{27.28.29.30})\)

\(=\frac{1}{3}(\frac{4}{1.2.3.4}-\frac{1}{1.2.3.4}+\frac{5}{2.3.4.5}-\frac{2}{2.3.4.5}+...+\frac{30}{27.28.29.30}-\frac{27}{27.28.29.30})\)

\(=\frac{1}{3}(\frac{1}{1.2.3}-\frac{1}{2.3.4}+\frac{1}{2.3.4}-\frac{1}{3.4.5}+...+\frac{1}{27.28.29}-\frac{1}{28.29.30})\)

\(=\frac{1}{3}\left(\frac{1}{1.2.3}-\frac{1}{28.29.30}\right)\)

\(=\frac{1}{3}\left(\frac{1}{6}-\frac{1}{24360}\right)\)

\(=\frac{1}{3}.\frac{1353}{8120}\)

\(=\frac{451}{8120}\)

kieu mo mau no the(dung hoi vi sao)?

1.2.3.

=>tiep theo la 4

Khi gặp dạng như thế này, ta xét số hạng như thế này thì ta sẽ có được số cần nhân chính là số liền sau của số cuối cùng trong tích đó. Nói dễ hiểu hơn là nếu có A = 1.2 + 2.3 + 3.4 +... thì ta xét số hạng đầu tiên của tổng là 1.2 thì ta có số liền sau của 2 là 3. Vậy nên nhân A cho 3. Cái này gọi là quy luật để giải quyết bài toán kiểu này rồi.

Ta có: B = 1.2.3 + 2.3.4 + 3.4.5 + ... + 17.18.19
=> 4B = 4(1.2.3 + 2.3.4 + 3.4.5 + ... + 17.18.19)
=> 4B = 1.2.3.4 + 2.3.4.4 + 3.4.5.4 +...... +17.18.19.4
=> 4B = 1.2.3.4 + 2.3.4(5 - 1) + 3.4.5.(6 - 2) +..... +17.18.19.(20 - 16)
=> 4B = 1.2.3.4 + 2.3.4.5 - 2.3.4 + 3.4.5.6 - 2.3.4.5 + ..... + 17.18.19.20 - 16.17.18.19
=> 4B = 17.18.19.20
=> 4B = 116280
=> B = 29070

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

=> a = \(\frac{1}{2}\left(\frac{1}{1}-\frac{1}{2}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{3}\right)\)+\(\frac{1}{2}\left(\frac{1}{2}-\frac{1}{3}\right)-\frac{1}{2}\left(\frac{1}{3}-\frac{1}{4}\right)\)+....+\(\frac{1}{2}\left(\frac{1}{2018}-\frac{1}{2019}\right)-\frac{1}{2}\left(\frac{1}{2019}-\frac{1}{2020}\right)\)=\(\frac{1}{2}\left(1-\frac{1}{2}\right)-\frac{1}{2}\left(\frac{1}{2019}-\frac{1}{2020}\right)\)=\(\frac{1}{4}\left(1-\frac{1}{2019.1010}\right)\)=\(\frac{2019.1010-1}{2.2019.2020}\)

b) tương tự \(\frac{1}{n\left(n+1\right)\left(n+2\right)\left(n+3\right)}=\left(\frac{1}{n}-\frac{1}{n+1}\right)\left(\frac{1}{n+2}-\frac{1}{n+3}\right)\)=\(\frac{1}{2}\left(\frac{1}{n}-\frac{1}{n+2}\right)-\left(\frac{1}{n+1}-\frac{1}{n+2}\right)\)-\(\frac{1}{3}\left(\frac{1}{n}-\frac{1}{n+3}\right)+\frac{1}{2}\left(\frac{1}{n+1}-\frac{1}{n+3}\right)\)=\(\frac{1}{6}\left(\frac{1}{n}-\frac{1}{n+1}\right)-\frac{1}{3}\left(\frac{1}{n+1}-\frac{1}{n+2}\right)\)+\(\frac{1}{6}\left(\frac{1}{n+2}-\frac{1}{n+3}\right)\)= M-P+N

Với n từ 1 đến 2017 thì

M= \(\frac{1}{6}\left(\frac{1}{1}-\frac{1}{2}\right)+\frac{1}{6}\left(\frac{1}{2}-\frac{1}{3}\right)+...\)+\(\frac{1}{6}\left(\frac{1}{2017}-\frac{1}{2018}\right)\)=\(\frac{1}{6}\left(1-\frac{1}{2018}\right)=\frac{2017}{6.2018}\)

N= \(\frac{1}{6}\left(\frac{1}{3}-\frac{1}{4}\right)+\frac{1}{6}\left(\frac{1}{4}-\frac{1}{5}\right)+...+\)\(\frac{1}{6}\left(\frac{1}{2019}-\frac{1}{2020}\right)=\)\(\frac{1}{6}\left(\frac{1}{3}-\frac{1}{2020}\right)=\frac{2017}{6.3.2020}\)

P= \(\frac{1}{3}\left(\frac{1}{2}-\frac{1}{3}\right)+\frac{1}{3}\left(\frac{1}{3}-\frac{1}{4}\right)+...+\)\(\frac{1}{3}\left(\frac{1}{2018}-\frac{1}{2019}\right)\)= \(\frac{1}{3}\left(\frac{1}{2}-\frac{1}{2019}\right)=\frac{2017}{3.2.2019}\)

M+N-P = \(\frac{2017}{6}\left(\frac{1}{2018}+\frac{1}{3.2020}-\frac{1}{2019}\right)\)=\(\frac{2017}{6}.\left(\frac{1}{2018.2019}+\frac{1}{3.2020}\right)\)

=  \(\frac{2017\left(1010+1009.673\right)}{3.2018.2019.2020}\)

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 \(A=\frac{3^{2008}-1}{2.3^{2008}}\)

a) \(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{27.28.29}\)

\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{27.28}-\frac{1}{28.29}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{28.29}\right)\)

\(=\frac{1}{2}.\frac{405}{812}=\frac{405}{1624}\)

Vậy giá trị của biểu thức \(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{27.28.29}=\frac{405}{1624}\)

b) \(\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+\frac{1}{3.4.5.6}+...+\frac{1}{27.28.29.30}\)

\(=\frac{1}{3}.\left(\frac{1}{1.2.3}-\frac{1}{2.3.4}+\frac{1}{2.3.4}-\frac{1}{3.4.5}+\frac{1}{3.4.5}-\frac{1}{4.5.6}+....+\frac{1}{27.28.29}-\frac{1}{28.29.30}\right)\)

\(=\frac{1}{3}.\left(\frac{1}{1.2.3}-\frac{1}{28.29.30}\right)\)

\(=\frac{1}{3}\cdot\frac{1353}{8120}=\frac{451}{8120}\)

Vậy giá trị của biểu thức \(\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+\frac{1}{3.4.5.6}+...+\frac{1}{27.28.29.30}=\frac{451}{8120}\)

\(A=\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+...+\frac{1}{n\left(n+1\right)\left(n+2\right)\left(n+3\right)}\)

\(3A=\frac{3}{1.2.3.4}+\frac{3}{2.3.4.5}+...+\frac{3}{n\left(n+1\right)\left(n+2\right)\left(n+3\right)}\)

\(3A=\frac{1}{1.2.3}-\frac{1}{2.3.4}+\frac{1}{2.3.4}-\frac{1}{3.4.5}+...+\frac{1}{n\left(n+1\right)\left(n+2\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)\left(n+3\right)}\)

\(3A=\frac{1}{6}-\frac{1}{\left(n+1\right)\left(n+2\right)\left(n+3\right)}\)

\(3A=\frac{\left(n+1\right)\left(n+2\right)\left(n+3\right)-6}{6\left(n+1\right)\left(n+2\right)\left(n+3\right)}\)

=>\(A=\frac{\left(n+1\right)\left(n+2\right)\left(n+3\right)-6}{18\left(n+1\right)\left(n+2\right)\left(n+3\right)}=\frac{n^3+3n^2+3n^2+9n+6-6}{18\left(n+1\right)\left(n+2\right)\left(n+3\right)}=\frac{n^3+6n^2+9n}{18\left(n+1\right)\left(n+2\right)\left(n+3\right)}\)