Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Trước tiên, chúng ta cần có lý thuyết về biến đổi phân số.
\(\dfrac{b-a}{a\cdot b}=\dfrac{1}{a}-\dfrac{1}{b}\)
Ta có:
\(S=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{2017\cdot2018}\)
\(S=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2017}-\dfrac{1}{2018}\)
\(S=1+\left(-\dfrac{1}{2}+\dfrac{1}{2}\right)+\left(-\dfrac{1}{3}+\dfrac{1}{3}\right)+...-\dfrac{1}{2018}\)
\(S=1-\dfrac{1}{2018}\)
\(S=\dfrac{2017}{2018}\)
=1/1.2+1/2.3+1/3.4+...1/2017.2018
=1/1-1/2+1/2-1/3+1/3-1/4+...+1/2017-1/2018
=1-1/2018
=2018/2018-1/2018
=2017/2018
\(B=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(B=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(B=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)\)
\(B=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)
\(B< \frac{1}{60}+\frac{1}{60}+...+\frac{1}{60}\)
\(B< \frac{50}{60}\Leftrightarrow B< \frac{5}{6}\)
\(\frac{1}{1.2}\)\(+\frac{1}{2.3}+\)\(\frac{1}{3.4}\)\(+\)\(.............+\)\(\frac{1}{2017.2018}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+......+\frac{1}{2017}-\frac{1}{2018}\)
\(=\frac{1}{1}-\frac{1}{2018}\)
\(=\frac{2017}{2018}\)
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.........+\frac{1}{2017.2018}\)
\(=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+......+\frac{2018-2017}{2017.2018}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{2017}-\frac{1}{2018}\)
\(=1-\frac{1}{2018}\)
\(=\frac{2017}{2018}\)
Ta có công thức :
\(\frac{1}{k\left(k+1\right)}=\frac{\left(k+1\right)-k}{k\left(k+1\right)}=\frac{k+1}{k\left(k+1\right)}-\frac{k}{k\left(k+1\right)}=\frac{1}{k}-\frac{1}{k+1}\)
\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{n-1}-\frac{1}{n}\)
\(=1-\frac{1}{n}=\frac{n-1}{n}\)
Ta có: A = 1.2 + 2.3 + 3.4 + … + n.(n + 1)
=> 3A = 1.2.(3-0) + 2.3.(4-1) + .... + n.(n+1).(n+2 - n+1)
=> 3A = 1.2.3 - 1.2.3 + 2.3.4 - 2.3.4 + .... + n.(n+1).(n+2)
=> 3A = n.(n+1).(n+2)
= > A = \(\frac{\text{n.(n+1).(n+2)}}{3}\)
Tổng quát:\(1-\frac{1}{1+2+......+n}=1-\frac{1}{\frac{n\left(n+1\right)}{2}}=1-\frac{2}{n\left(n+1\right)}=\frac{n^2+n-2}{n\left(n+1\right)}\)
\(=\frac{n^2-n+2n-2}{n\left(n+1\right)}=\frac{n\left(n-1\right)+2\left(n-1\right)}{n\left(n+1\right)}=\frac{\left(n+2\right)\left(n-1\right)}{n\left(n+1\right)}\) với \(n\in\)N*
Thay x=2,x=3,..........,x=2018 vào ta có:
\(\left(1-\frac{1}{1+2}\right)\left(1-\frac{1}{1+2+3}\right)......\left(1-\frac{1}{1+2+3+.....+2018}\right)=\frac{1.4}{2.3}.\frac{2.5}{3.4}.........\frac{2017.2020}{2018.2019}\)
\(=\frac{1.2.3......2017}{2.3.......2018}.\frac{4.5........2020}{3.4.......2019}=\frac{1}{2018}.\frac{2020}{3}=\frac{2020}{6054}=\frac{1010}{3027}\)
=1(1+1) + 2(2+1) + 3(3+1) +...+n(n+1)
=(1^2 + 2^2 + 3^2 +...+ n^2) + (1 + 2 + 3 + ...+ n)
ta có các công thức:
1^2 + 2^2 + 3^2 +...+ n^2 = n(n+1)(2n+1)/6
1 + 2 + 3 + ...+ n = n(n+1)/2
thay vào ta có:
S = n(n+1)(2n+1)/6 + n(n+1)/2
=n(n+1)/2[(2n+1)/3 + 1]
=n(n+1)(n+2)/3
Siêu tốc tổng quát: \(\frac{1}{n.\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}\)áp vào
\(A=\frac{1}{1}-\frac{1}{14}=1-\frac{1}{14}\)
A=(2-1)/1.2+(3-2)/2.3+...+(14-13)/13.14
A=1-1/2+1/2-1/3+...+1/13-1/14
A=1-1/14=13/14