Ta có: \(S=\frac{1}{2^2}-\frac{1}{2^4}+\frac{1}{2^6}-\frac{1}{2^8}+\cdots+\frac{1}{2^{2002}}-\frac{1}{2^{2004}}\)

=>4S=\(1-\frac{1}{2^2}+\frac{1}{2^4}-\frac{1}{2^6}+\cdots+\frac{1}{2^{2000}}-\frac{1}{2^{2002}}\)

=>4S+S=\(1-\frac{1}{2^2}+\frac{1}{2^4}-\frac{1}{2^6}+\cdots+\frac{1}{2^{2000}}-\frac{1}{2^{2002}}+1-\frac{1}{2^2}+\frac{1}{2^4}-\frac{1}{2^6}+\cdots+\frac{1}{2^{2000}}-\frac{1}{2^{2002}}\)

=>5S=\(1-\frac{1}{2^{2002}}\)

=>5S<1

=>S<0,2

1 - 1/2 + 1/3 - 1/4 + .. + 1/121 - 1/122 + 1/123

=  ( 1 + 1/3 + ... + 1/121 + 1/123 ) - ( 1/2 + 1/4 + .. + 1/122 ) 

Đến bước này ta sẽ cùng cộng 2 vế với : 1/2 + 1/4 + .. + 1/122 

=   ( 1 + 1/2 + 1/3 + 1/4 + ...+ 1/121 + 1/122 + 1/123 ) - 2. ( 1/2 + 1/4 + .. + 1/122 )

=    (  1 + 1/2 + 1/3 + ...+ 1/122 +1 /123 )  -  (  1 + 1/2 + ...+ 1/61 ) 

=                   1/62 + 1/63 + ..+1/122 + 1/123 

Chúc học giỏi !!

Ta có: \(\frac{1}{2^2}<\frac{1}{1\cdot2}=1-\frac12\)

\(\frac{1}{3^2}<\frac{1}{2\cdot3}=\frac12-\frac13\)

...

\(\frac{1}{n^2}<\frac{1}{\left(n-1\right)\cdot n}=\frac{1}{n-1}-\frac{1}{n}\)

Do đó: \(\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{n^2}<1-\frac12+\frac12-\frac13+\cdots+\frac{1}{n-1}-\frac{1}{n}\)

=>\(\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{n^2}<1-\frac{1}{n}<1\)

=>\(1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{n^2}<1+1=2\)

=>A<2

Ta có: \(\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{n^2}>0\)

=>\(1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{n^2}>1\)

=>A>1

Do đó: 1<A<2

=>A không là số tự nhiên

\(\frac{121+122}{2}\)=\(\frac{243}{2}\)=243:2=121,5

(121+122):2

=   243     :2

=121, 5.

244

nha

122+122=244

122+122=244

122+122=244