1/1×2 + 1/2×3 + 1/3×4 + 1/4×5 + ... + 1/99×100

= 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + ... + 1/99 - 1/100

= 1 - 1/100

= 99/100

\(\text{S}\)= 1 - \(\frac{1}{2}\)+ \(\frac{1}{2}\)-\(\frac{1}{3}\)+ \(\frac{1}{3}\)- \(\frac{1}{4}\)+ .... + \(\frac{1}{99}\)- \(\frac{1}{100}\)

\(S\)= ( 1 - \(\frac{1}{100}\)) : 2

\(S\)= \(\frac{99}{100}\): 2 

\(S\)= \(\frac{99}{200}\)

tick nhé Lê Thiên Hương

99/200 dạng chuỗi mà bạn

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}=1-\frac{1}{100}=\frac{99}{100}\)

Đặt \(M=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)

\(M=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)

\(M=1-\frac{1}{100}\)

\(M=\frac{99}{100}\)

\(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+....+\frac{1}{99\times100}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

\(=1-\frac{1}{100}\)

\(=\frac{99}{100}\)

\(S=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{99\cdot100}\)

Áp dụng công thức : \(\frac{1}{n\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}\)

\(S=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)

\(S=1-\frac{1}{100}=\frac{99}{100}\)

Dap an la 99/100.nho k cho minh.bai giai se gui sau

= 1/9900

đúng đấy

\(\frac{3}{1.2}+\frac{3}{2.3}+........+\frac{3}{99.100}\)

\(=3\left(\frac{1}{1.2}+\frac{1}{2.3}+.....+\frac{1}{99.100}\right)\)

\(=3\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.........+\frac{1}{99}-\frac{1}{100}\right)\)

\(=3\left(1-\frac{1}{100}\right)\)

\(=\frac{3.99}{100}=\frac{297}{100}\)

=1/1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+...+1/98-1/99+1/99-1/100

=1/1-1/100

=100/100-1/100

=99/100

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

= \(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

= \(\frac{1}{1}-\frac{1}{100}\)

= \(\frac{99}{100}\)

~~~
#Sunrise

Đặt \(A=\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+....+\frac{1}{99x100}\)

\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{99}-\frac{1}{100}\)

\(\Rightarrow A=1-\frac{1}{100}\)

\(\Rightarrow A=\frac{99}{100}\)

Hay \(\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+....+\frac{1}{99x100}=\frac{99}{100}\)

\(M=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)

\(\Rightarrow M=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)

\(\Rightarrow M=1-\frac{1}{100}\)

\(\Rightarrow M=\frac{100}{100}-\frac{1}{100}=\frac{99}{100}\)

\(b,N=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\)

\(\Rightarrow N=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\right)\)

\(\Rightarrow N=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+..+\frac{1}{97}-\frac{1}{99}\right)\)

\(\Rightarrow N=\frac{1}{2}.\left(1-\frac{1}{99}\right)=\frac{1}{2}.\frac{98}{99}\)

\(\Rightarrow N=\frac{1.98}{2.99}=\frac{49.2}{2.99}=\frac{49}{99}\)

\(a,M=1-\frac{1}{100}=\frac{99}{100}\)

\(b=2N=\frac{2}{1x3}+\frac{2}{3x5}+\frac{2}{5x7}+...+\frac{2}{97x99}\)

                  \(=1-\frac{1}{99}=\frac{98}{99}\)

   =>\(N=\frac{98}{99}:2=\frac{49}{99}\)

\(M=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)

\(M=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)

\(M=1-\frac{1}{100}\)

\(M=\frac{99}{100}\)

là 99/100

Ta có :\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+......+\frac{1}{99.100}\) 

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+......+\frac{1}{99}-\frac{1}{100}\)

\(=1-\frac{1}{100}\)

\(=\frac{99}{100}\)

\(\frac{1}{1\times2}+...+\frac{1}{99\times100}\)

= \(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...+\frac{1}{99}-\frac{1}{100}\)

= \(\frac{1}{1}-\frac{1}{100}=\frac{99}{100}\)

\(B=1+\frac{1}{1.2}+\frac{1}{2.3}+.....+\frac{1}{99.100}.\)

\(B=1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+........+\frac{1}{99}+\frac{1}{100}\)

\(B=1+1-\frac{1}{100}=2-\frac{1}{100}\)

\(B=\frac{199}{100}\)

\(C=\frac{1}{1.2}+\frac{1}{2.3}+........+\frac{1}{n\left(n+1\right)}\)

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

\(C=1-\frac{1}{n+1}\)

\(C=\frac{n+1-1}{n+1}=\frac{n}{n+1}\)

Áp dụng công thức tình dãy số ta có :

\(D=\frac{\left[\left(n-1\right):1+1\right].\left(n+1\right)}{2}=\frac{n.\left(n+1\right)}{2}\)

cảm ơn chị nhiều lắm ạ !