a:

Số số hạng trong dãy M là:

(1002-12):10+1=100(số)

=>Sẽ có 50 cặp (1002;992); (982;972);....;(22;12) có hiệu bằng 10

\(M=1002-992+982-972+...+22-12\)

\(=\left(1002-992\right)+\left(982-972\right)+...+\left(22-12\right)\)

\(=10+10+...+10\)

=10*50=500

b: \(N=\left(202+182+...+42+22\right)-\left(192+172+...+32+12\right)\)

\(=\left(202-192\right)+\left(182-172\right)+...+\left(22-12\right)\)

=10+10+...+10

=10*10=100

a) \(=\left(127+73\right)^2=200^2=40000\)

b) \(=18^8-\left(18^8-1\right)=1\)

c) \(=\left(100+99\right)\left(100-99\right)+\left(98+97\right)\left(98-97\right)+...+\left(2+1\right)\left(2-1\right)\)

\(=100+99+98+97+...+2+1=5050\)

d) biến đổi thành \(20^2-19^2+18^2-17^2+..+2^2-1^2\)

rồi giải ra như trên

Đặt A=\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{100^2}\)

Ta có: \(\dfrac{1}{2^2}< \dfrac{1}{1.2},\dfrac{1}{3^2}< \dfrac{1}{2.3},...,\dfrac{1}{100^2}< \dfrac{1}{99.100}\)

\(A\)<\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\)

A<\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

A<\(1-\dfrac{1}{100}=\dfrac{99}{100}\)(đpcm)

Ta có: \(\dfrac{1}{2^2}>\dfrac{1}{2.3},\dfrac{1}{3^2}>\dfrac{1}{3.4},...,\dfrac{1}{100^2}>\dfrac{1}{100.101}\)

A>\(\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{100.101}\)

A>\(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{100}-\dfrac{1}{101}\)

A>\(\dfrac{1}{2}-\dfrac{1}{101}=\dfrac{99}{202}\)(đpcm)

Vậy \(\dfrac{99}{100}>A>\dfrac{99}{202}\)

fan bé sans à

wuttttt

Đặt \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\) ta có : 

\(\frac{1}{2^2}>\frac{1}{2.3}\)

\(\frac{1}{3^2}>\frac{1}{3.4}\)

\(\frac{1}{4^2}>\frac{1}{4.5}\)

\(............\)

\(\frac{1}{100^2}>\frac{1}{100.101}\)

\(\Rightarrow\)\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}>\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{100.101}\)

\(\Rightarrow\)\(A>\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{100}-\frac{1}{101}\)

\(\Rightarrow\)\(A>\frac{1}{2}-\frac{1}{101}\)

\(\Rightarrow\)\(A>\frac{99}{202}\) \(\left(1\right)\)

Lại có : 

\(\frac{1}{2^2}< \frac{1}{1.2}\)

\(\frac{1}{3^2}< \frac{1}{2.3}\)

\(\frac{1}{4^2}< \frac{1}{3.4}\)

\(............\)

\(\frac{1}{100^2}< \frac{1}{99.100}\)

\(\Rightarrow\)\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

\(\Rightarrow\)\(A< \frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

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

\(\Rightarrow\)\(A< \frac{99}{100}\) \(\left(2\right)\)

Từ (1) và (2) suy ra : \(\frac{99}{202}< A< \frac{99}{100}\) ( đpcm ) 

Vậy \(\frac{99}{202}< A< \frac{99}{100}\)

Chúc bạn học tốt ~ 

Ta có \(\frac{1}{1945^2}+\frac{1}{1946^2}+\frac{1}{1947^2}+...+\frac{1}{1975^2}\)

\(< \frac{1}{1944\cdot1945}+\frac{1}{1945\cdot1946}+...+\frac{1}{1974.1975}\)

\(=\frac{1}{1944}-\frac{1}{1945}+\frac{1}{1945}-\frac{1}{1946}+...+\frac{1}{1974}-\frac{1}{1975}\)

=\(\frac{1}{1944}-\frac{1}{1975}< \frac{1}{1944}\)

\(\Rightarrow\frac{1}{1945^2}+\frac{1}{1946^2}+\frac{1}{1947^2}+..+\frac{1}{1975^2}< \frac{1}{1944}\)

╔♫═╗╔╗ ♥
╚╗╔╝║║♫═╦╦╦╔╗
╔╝╚╗♫╚╣║║║║╔╣
╚═♫╝╚═╩═╩♫╩═╝ ஜ۩۞۩ஜ YOU ஜ۩۞۩ஜ