\(S=\dfrac{1}{2^2}-\dfrac{1}{2^4}+...+\dfrac{1}{2^{4n-2}}-\dfrac{1}{2^{4n}}+...+\dfrac{1}{2^{2002}}-\dfrac{1}{2^{2004}}\)

\(2^2S=2^2\left(\dfrac{1}{2^2}-\dfrac{1}{2^4}+...+\dfrac{1}{2^{4n-2}}-\dfrac{1}{2^{4n}}+...+\dfrac{1}{2^{2002}}-\dfrac{1}{2^{2004}}\right)\)

\(4S=1-\dfrac{1}{2^2}+...+\dfrac{1}{2^{4n}}+\dfrac{1}{2^{4n+2}}+...+\dfrac{1}{2^{2000}}-\dfrac{1}{2^{2002}}\)

\(4S+S=\left(1-\dfrac{1}{2^2}+...+\dfrac{1}{2^{2000}}-\dfrac{1}{2^{2002}}\right)+\left(\dfrac{1}{2^2}-\dfrac{1}{2^4}+...+\dfrac{1}{2^{2002}}-\dfrac{1}{2^{2004}}\right)\)

\(5S=1-\dfrac{1}{2^{2004}}< 1\Rightarrow S< \dfrac{1}{5}=0,2\)

Đặt :

\(A=\dfrac{1}{2^2}-\dfrac{1}{2^4}+\dfrac{1}{2^6}-......+\dfrac{1}{2^{4n-2}}-\dfrac{1}{2^{4n}}+.........+\dfrac{1}{2^{2002}}-\dfrac{1}{2^{2004}}\)

\(\Leftrightarrow2^2A=2^2\left(\dfrac{1}{2^2}-\dfrac{1}{2^4}+\dfrac{1}{2^6}-.......+\dfrac{1}{2^{4n-2}}-\dfrac{1}{2^{4n}}+......+\dfrac{1}{2^{2002}}-\dfrac{1}{2^{2004}}\right)\)

\(\Leftrightarrow4A=1-\dfrac{1}{2^2}+\dfrac{1}{2^4}-\dfrac{1}{2^6}+.......-\dfrac{1}{2^{4n-2}}+\dfrac{1}{2^{4n}}-.......-\dfrac{1}{2^{2002}}\)

\(\Leftrightarrow4A+A=\left(1-\dfrac{1}{2^2}+\dfrac{1}{2^4}-\dfrac{1}{2^6}+.......-\dfrac{1}{2^{4n-2}}+\dfrac{1}{2^{4n}}-......-\dfrac{1}{2^{2002}}\right)+\left(\dfrac{1}{2^2}-\dfrac{1}{2^4}+......+\dfrac{1}{2^{4n-2}}-\dfrac{1}{2^{4n}}+......+\dfrac{1}{2^{2002}}-\dfrac{1}{2^{2004}}\right)\)

\(\Leftrightarrow5A=1-\dfrac{1}{2^{2004}}\)

\(\Leftrightarrow A=\left(1-\dfrac{1}{2^{2004}}\right):5\)

\(\Leftrightarrow A=\dfrac{1}{5}-\dfrac{1}{5}.\dfrac{1}{2^{2004}}< \dfrac{1}{5}=0,2\left(đpcm\right)\)

a)

\(B=1-\dfrac{1}{2^2}-\dfrac{1}{3^2}-\dfrac{1}{4^2}-...........-\dfrac{1}{2004^2}\)

\(\Leftrightarrow B=1-\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+..............+\dfrac{1}{2004^2}\right)\)

Đặt :

\(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+.............+\dfrac{1}{2004^2}\)

Ta thấy :

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

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

..........................

\(\dfrac{1}{2004^2}< \dfrac{1}{2003.2004}\)

\(\Leftrightarrow A< \dfrac{1}{1.2}+\dfrac{1}{2.3}+..............+\dfrac{1}{2003.2004}\)

\(\Leftrightarrow A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+..........+\dfrac{1}{2003}-\dfrac{1}{2004}\)

\(\Leftrightarrow A< 1-\dfrac{1}{2004}\)

\(\Leftrightarrow A< \dfrac{2003}{2004}\)

\(\Leftrightarrow1-A< 1-\dfrac{2003}{2004}\)

\(\Leftrightarrow B< \dfrac{1}{2004}\left(đpcm\right)\)

b) \(S=\dfrac{1}{2^2}-\dfrac{1}{2^4}+\dfrac{1}{2^6}-........+\dfrac{1}{2^{4n-2}}-\dfrac{1}{2^{4n}}+.......+\dfrac{1}{2^{2002}}-\dfrac{1}{2^{2004}}\)

\(\Leftrightarrow2^2S=2^2\left(\dfrac{1}{2^2}-\dfrac{1}{2^4}+.....+\dfrac{1}{2^{4n-2}}-\dfrac{1}{2^{4n}}+....+\dfrac{1}{2^{2002}}-\dfrac{1}{2^{2004}}\right)\)

\(\Leftrightarrow4S=1-\dfrac{1}{2^2}+.......+\dfrac{1}{2^{4n}}-\dfrac{1}{2^{4n+2}}+.......+\dfrac{1}{2^{2000}}-\dfrac{1}{2^{2002}}\)

\(\Leftrightarrow4S+S=\left(1-\dfrac{1}{2^2}+.....+\dfrac{1}{2^{2000}}-\dfrac{1}{2^{2002}}\right)+\left(\dfrac{1}{2^2}-\dfrac{1}{2^4}+.......+\dfrac{1}{2^{2002}}-\dfrac{1}{2^{2004}}\right)\)\(\Leftrightarrow5S=1-\dfrac{1}{2^{2004}}< 1\)

\(\Leftrightarrow S< \dfrac{1}{5}=0,2\)

\(\Leftrightarrow S< 0,2\left(đpcm\right)\)

\(S=\dfrac{1}{2^2}-\dfrac{1}{2^4}+\dfrac{1}{2^6}-......+\dfrac{1}{2^{4n-2}}-\dfrac{1}{2^{4n}}+......+\dfrac{1}{2^{2002}}-\dfrac{1}{2^{2004}}\Rightarrow4S=1-\dfrac{1}{2^2}+\dfrac{1}{2^4}-\dfrac{1}{2^6}+......-\dfrac{1}{2^{4n-2}}+\dfrac{1}{2^{4n}}+......-\dfrac{1}{2^{2002}}\Rightarrow4S+S=5S=1-\dfrac{1}{2^{2004}}< 1\Rightarrow S< 0,2\left(\text{đpcm}\right)\)

A=\(\dfrac{7^2-1}{7^4}+\dfrac{7^2-1}{7^8}+...+\dfrac{7^2-1}{7^{100}}=\left(7^2-1\right)\left(\dfrac{1}{7^4}+\dfrac{1}{7^8}+...+\dfrac{1}{7^{100}}\right)=48\cdot B\)Dễ dàng tính được B( nhân hết với 7 mũ 4 roi trừ đi, chia ra là xong) ra đpcm.

Lên lớp 11 thì ta có dạng tổng quát luôn này(tức là nếu n quá lớn thì có thể coi là xảy ra dấu bằng) \(\dfrac{1}{7^2}-\dfrac{1}{7^4}+...+\dfrac{1}{7^n}-\dfrac{1}{7^{n+2}}< \dfrac{1}{50}\)

Lời giải:

Đặt \(A=\frac{1}{7^2}-\frac{1}{7^4}+....+\frac{1}{7^{4n-2}}-\frac{1}{7^{4n}}+...+\frac{1}{7^{98}}-\frac{1}{7^{100}}\)

\(7^2A=1-\frac{1}{7^2}+....+\frac{1}{7^{4n-4}}-\frac{1}{7^{4n-2}}+...+\frac{1}{7^{96}}-\frac{1}{7^{98}}\)

\(\Rightarrow A+7^2A=1-\frac{1}{7^{100}}\Rightarrow 50A=1-\frac{1}{7^{100}}<1\)

$\Rightarrow A< \frac{1}{50}$

Đặt \(S=\dfrac{1}{7^2}-\dfrac{1}{7^4}+...+\dfrac{1}{7^{4n-2}}-\dfrac{1}{7^{4n}}+...+\dfrac{1}{7^{98}}-\dfrac{1}{7^{100}}\)

\(\Rightarrow\dfrac{S}{7^2}=\dfrac{1}{7^4}-\dfrac{1}{7^6}+...+\dfrac{1}{7^{100}}-\dfrac{1}{7^{102}}\)

\(\Rightarrow S+\dfrac{S}{7^2}=\left(\dfrac{1}{7^2}-\dfrac{1}{7^4}+...+\dfrac{1}{7^{98}}-\dfrac{1}{7^{100}}\right)+\left(\dfrac{1}{7^4}-\dfrac{1}{7^6}+...+\dfrac{1}{7^{100}}-\dfrac{1}{7^{102}}\right)\)

\(\Leftrightarrow\dfrac{50S}{49}=\dfrac{1}{7^2}-\dfrac{1}{7^{102}}< \dfrac{1}{7^2}=\dfrac{1}{49}< \dfrac{1}{50}\)

\(\Leftrightarrow S< \dfrac{1}{50}\)

Vậy \(\dfrac{1}{7^2}-\dfrac{1}{7^4}+...+\dfrac{1}{7^{98}}-\dfrac{1}{7^{100}}< \dfrac{1}{50}\) (Đpcm)

Đặt \(A=\dfrac{1}{7^2}-\dfrac{1}{7^4}+...+\dfrac{1}{7^{4n-2}}-\dfrac{1}{7^{4n}}+...+\dfrac{1}{7^{98}}+\dfrac{1}{7^{100}}\)

Ta có:

\(\dfrac{A}{7^2}=\dfrac{1}{7^4}-\dfrac{1}{7^6}+...+\dfrac{1}{7^{100}}+\dfrac{1}{7^{102}}\)

\(\Rightarrow A+\dfrac{A}{7^2}=\left(\dfrac{1}{7^2}-\dfrac{1}{7^4}+...+\dfrac{1}{7^{98}}+\dfrac{1}{7^{100}}\right)+\left(\dfrac{1}{7^4}-\dfrac{1}{7^6}+...+\dfrac{1}{7^{100}}+\dfrac{1}{7^{102}}\right)\)

\(\Rightarrow\dfrac{50A}{49}=\dfrac{1}{7^2}-\dfrac{1}{7^{102}}< \dfrac{1}{7^2}=\dfrac{1}{49}\)

\(\Rightarrow A< \dfrac{1}{50}\)

=> ĐPCM.