Đặt \(A=\dfrac{1}{4}+\dfrac{1}{16}+\dfrac{1}{36}+\dfrac{1}{64}+...+\dfrac{1}{10000}\)

Ta có:

\(A=\dfrac{1}{4}+\dfrac{1}{16}+\dfrac{1}{36}+\dfrac{1}{64}+...+\dfrac{1}{10000}\)

\(\Rightarrow A=\dfrac{1}{4}\left(1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{50^2}\right)\)

\(\Rightarrow A< \dfrac{1}{4}\left(1+\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{49.50}\right)\)

\(\Rightarrow A< \dfrac{1}{4}\left(1+1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)\)

\(\Rightarrow A< \dfrac{1}{4}\left(1+1-\dfrac{1}{50}\right)\)

\(\Rightarrow A< \dfrac{1}{4}.\dfrac{99}{50}\)

\(\Rightarrow A< \dfrac{99}{200}< \dfrac{1}{2}\)

Vậy \(\dfrac{1}{4}+\dfrac{1}{16}+\dfrac{1}{36}+\dfrac{1}{64}+...+\dfrac{1}{10000}< \dfrac{1}{2}\) (Đpcm)

\(\dfrac{1}{4}+\dfrac{1}{16}+\dfrac{1}{36}+...+\dfrac{1}{10000}=\dfrac{1}{2^2}+\dfrac{1}{4^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}\)

\(=\dfrac{1}{2^2}\left(1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{50^2}\right)< \dfrac{1}{4}\left(1+\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}\right)=\dfrac{1}{4}\left(1+1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)=\dfrac{1}{4}\left(1+1-\dfrac{1}{50}\right)=\dfrac{1}{4}\left(2-\dfrac{1}{50}\right)< \dfrac{1}{2}\)

\(\Rightarrow\dfrac{1}{4}+\dfrac{1}{16}+\dfrac{1}{36}+...+\dfrac{1}{10000}< \dfrac{1}{2}\)

\(A=\dfrac{1}{4}\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{1010^2}\right)\)

1/2^2+1/3^2+...+1/2010^2<1/1*2+1/2*3+...+1/2009*2010=1-1/2010<1

=>A<1/4

Ta có : \(\dfrac{1}{4}\)= \(\dfrac{1}{2.2}\)> \(\dfrac{1}{2.3}\)

\(\dfrac{1}{9}\)= \(\dfrac{1}{3.3}\)> \(\dfrac{1}{3.4}\)

\(\dfrac{1}{16}\)=\(\dfrac{1}{4.4}\)> \(\dfrac{1}{4.5}\)

.......

\(\dfrac{1}{9801}\)= \(\dfrac{1}{99.99}\)> \(\dfrac{1}{99.100}\)

\(\dfrac{1}{10000}\)= \(\dfrac{1}{100.100}\)> \(\dfrac{1}{100.101}\)

\(\Rightarrow\) \(\dfrac{1}{4}\)+ \(\dfrac{1}{9}\)+ \(\dfrac{1}{16}\)+ ..... + \(\dfrac{1}{9801}\)+ \(\dfrac{1}{10000}\)> \(\dfrac{1}{2.3}\)+ \(\dfrac{1}{3.4}\)+ \(\dfrac{1}{4.5}\)+...+ \(\dfrac{1}{99.100}\)+\(\dfrac{1}{100.101}\)

= \(\dfrac{3-2}{2.3}\)+ \(\dfrac{4-3}{3.4}\)+ \(\dfrac{5-4}{4.5}\) +...+ \(\dfrac{100-99}{99.100}\)+ \(\dfrac{101-100}{100.101}\)

= \(\dfrac{3}{2.3}\)- \(\dfrac{2}{2.3}\) + \(\dfrac{4}{3.4}\)-\(\dfrac{3}{3.4}\)+ \(\dfrac{5}{4.5}\)-\(\dfrac{4}{4.5}\)+...+ \(\dfrac{100}{99.100}\)- \(\dfrac{99}{99.100}\)+ \(\dfrac{101}{100.101}\)-\(\dfrac{100}{100.101}\)

= \(\dfrac{1}{2}\)-\(\dfrac{1}{3}\)+\(\dfrac{1}{3}\)-\(\dfrac{1}{4}\)+ \(\dfrac{1}{4}\)-\(\dfrac{1}{5}\)+....+ \(\dfrac{1}{99}\)-\(\dfrac{1}{100}\)+\(\dfrac{1}{100}\)-\(\dfrac{1}{101}\)

= \(\dfrac{1}{2}\)- \(\dfrac{1}{101}\) ; Mà \(\dfrac{1}{2}\)- \(\dfrac{1}{101}\)= \(\dfrac{99}{202}\)< \(\dfrac{1}{2}\)

\(\Rightarrow\) \(\dfrac{1}{2}\)< \(\dfrac{1}{4}\)+ \(\dfrac{1}{9}\)+ \(\dfrac{1}{16}\)+...+ \(\dfrac{1}{9801}\)+ \(\dfrac{1}{10000}\) (1)

Ta có: \(VT=\dfrac{1}{2^2}+\dfrac{1}{4^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}\)

\(4VT=\dfrac{1}{2^2:2^2}+\dfrac{1}{4^2:2^2}+\dfrac{1}{6^2:2^2}+...+\dfrac{1}{100^2:2^2}\)

\(4VT=\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{50^2}\)

Lại có: \(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)

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

\(...\)

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

\(\Rightarrow4VT-1< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{49.50}\)(*)

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

\(=1-\dfrac{1}{50}\) (**)

Từ (*) và (**) \(\Rightarrow4VT< 2-\dfrac{1}{50}\)

\(\Rightarrow VT< \dfrac{1}{2}-\dfrac{1}{200}< VP\Rightarrow\) đpcm

b) Ta có: \(2VT=1-\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{8}+\dfrac{1}{16}-\dfrac{1}{32}\)

\(2VT+VT=\left(1-\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{8}+\dfrac{1}{16}-\dfrac{1}{32}\right)+\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+\dfrac{1}{32}-\dfrac{1}{64}\right)\)

\(3VT=1-\dfrac{1}{64}< 1\)

\(\Rightarrow VT< \dfrac{1}{3}\) (đpcm)

Giải thích các bước giải:

Đặt A= 1/4+1/16+1/36+1/64+1/100+1/144+1/196

= 1/2^2+ 1/4^2+ 1/6^2+….+ 1/16^2

= 1/2^2.( 1+ 1/2^2+ 1/3^2+…+ 1/8^2)

Ta có 1+ 1/2^2+ 1/3^2+…+ 1/8^2< 1+ 1/1.2+ 1/2.3+….7.8= 1+ 1-1/2+ 1/2- 1/3+….+ 1/7- 1/8

= 2- 1/8< 2

Nên ( 1+ 1/2^2+ 1/3^2+…+ 1/8^2)< 2

=> A< 1/2^2 nhân 2= 1/2

Vậy A< 1/2

Ta có:

\(\dfrac{n^2-1}{n^2}=1-\dfrac{1}{n^2}>1-\dfrac{1}{\left(n-1\right)n}\)

Áp dụng:

\(C=\dfrac{2^2-1}{2^2}+\dfrac{3^2-1}{3^2}+\dfrac{4^2-1}{4^2}+...+\dfrac{100^2-1}{100^2}\)

\(C>1-\dfrac{1}{1.2}+1-\dfrac{1}{2.3}+1-\dfrac{1}{3.4}+...+1-\dfrac{1}{99.100}\)

\(C>99-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\right)\)

\(C>99-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)

\(C>99-\left(1-\dfrac{1}{100}\right)\)

\(C>98+\dfrac{1}{100}>98\) (đpcm)

Theo bài ra, ta có:

\(S=\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{16}\)

\(\Rightarrow S=\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}\right)+\left(\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}\right)+\left(\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{11}\right)+\left(\dfrac{1}{12}+\dfrac{1}{13}+\dfrac{1}{14}\right)+\left(\dfrac{1}{15}+\dfrac{1}{16}\right)\)

\(\Rightarrow S< \left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}\right)+\dfrac{1}{6}.3+\dfrac{1}{9}.3+\dfrac{1}{12}.3+\dfrac{1}{15}.3\)

\(\Rightarrow S< \left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}\right)+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}\)

\(\Rightarrow S< 2\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}\right)\)

\(\Rightarrow S< 2\left[\left(\dfrac{1}{2}+\dfrac{1}{2}\right)+\left(\dfrac{1}{4}+\dfrac{1}{4}\right)\right]\)

\(\Rightarrow S< 2\left(\dfrac{2}{2}+\dfrac{2}{4}\right)\)

\(\Rightarrow S< 2\left(\dfrac{2}{2}+\dfrac{1}{2}\right)\)

\(\Rightarrow S< 2.\dfrac{3}{2}\)

\(\Rightarrow S< 3\left(1\right)\)

Lại có: \(S=\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{16}\)

\(\Rightarrow S=\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}\right)+\left(\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}\right)+\left(\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}\right)+\left(\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}+\dfrac{1}{16}\right)\)

\(\Rightarrow S>\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}\right)+\dfrac{1}{8}.4+\dfrac{1}{12}.4+\dfrac{1}{16}.4\)

\(\Rightarrow S>\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}\right)+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}\)

\(\Rightarrow S>2\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}\right)\)

\(\Rightarrow S>2\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{4}\right)\)

\(\Rightarrow S>2\left(\dfrac{1}{2}+\dfrac{2}{4}\right)\)

\(\Rightarrow S>2\left(\dfrac{1}{2}+\dfrac{1}{2}\right)\)

\(\Rightarrow S>2\)

Từ (1) và (2) suy ra \(2< S< 3\)

⇒ S không phải 1 số nguyên

Vậy...

Ta có \(\dfrac{6}{15}>\dfrac{6}{16}>...>\dfrac{6}{19}\) nên \(S< \dfrac{6}{15}.5=2\).

Lại có \(S>\dfrac{6}{19}.5>1\) nên \(1< S< 2\)

Ta có S = \(\dfrac{1}{31}+\dfrac{1}{32}+\dfrac{1}{33}+...+\dfrac{1}{60}=\left(\dfrac{1}{31}+\dfrac{1}{32}+...+\dfrac{1}{40}\right)+\left(\dfrac{1}{41}+\dfrac{1}{42}+...+\dfrac{1}{50}\right)+\left(\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{60}\right)\)⇒ S < \(\dfrac{1}{30}\cdot10+\dfrac{1}{40}\cdot10+\dfrac{1}{50}\cdot10=\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}=\dfrac{47}{60}< \dfrac{48}{60}=\dfrac{4}{5}\)

Vậy S < \(\dfrac{4}{5}\)