\(\text{S}\)= 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

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 !

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

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

k cho mình nha bạn

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

=1-1/100=99/100

I don't know

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

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

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

\(A=\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{99.100}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+..+\frac{1}{99}-\frac{1}{100}\)

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

\(=\left(1+\frac{1}{2}+...+\frac{1}{100}\right)-\left(1+\frac{1}{2}+...+\frac{1}{50}\right)=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)

Do \(\frac{1}{51}>\frac{1}{52}>...>\frac{1}{100}\Rightarrow A=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}>25\cdot\frac{1}{80}+25\cdot\frac{1}{100}=\frac{7}{12}\)

và \(A

olm lag kinh đang làm lag thoát ra mất tiêu

-------đề đúng------------

Ta có : 

Vậy ĐPCM

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

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

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

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

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

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

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

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

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

1/1x2+1/2x3+...+1/49x50

=1-1/2+1/2-1/3+.....+1/49-1/50

=1-1/50(1)

Ta co   1(2)

So sanh (1) voi (2) ta thay 1-1/50<1

=>1/1x2+...+1/49x50<1

(Phuong phap khu)

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{49.50}\)

=\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{49}-\frac{1}{50}\)

=\(\frac{1}{1}-\frac{1}{50}=\frac{50}{50}-\frac{1}{50}=\frac{49}{50}<1\)

Vậy \(\frac{49}{50}<1\)