hxbdsajfvgfchgcvetrfgeffbaw4bcdxrfbwebctfdefvewbdrgbwdgberhgcvergcdfbrbcftvvtgcbeftbckberfdkjuebfcbtuvrvbtkbr

Hoe..>>

Bài này mk gặp rồi nhờ cô giải hộ mà giờ mk quên mất tiêu rồi

Xin lỗi bn nha, mk k thể giúp đc rồi!

\(S=\frac{1}{2^2}-\frac{1}{2^4}+\frac{1}{2^6}+...+\frac{1}{2^{4n-2}}-\frac{1}{2^{4n}}+...\frac{1}{2^{2001}}-\frac{1}{2^{2004}}< 0.2\)

\(S=\left(\frac{1}{2.4}+\frac{1}{2.6}+\frac{1}{6.8}+\frac{1}{\left(2n-2\right).2n}\right).\frac{1}{2}\)

\(S=\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}+\frac{1}{6}+....\right)+\frac{1}{2n-2}-\frac{1}{2n}.\frac{1}{2}\)

\(S=\left(\frac{1}{2}-\frac{1}{2n}\right).\frac{1}{2}\)

\(S=\frac{1}{4}-\frac{1}{2n}< 0,2\)

\(S=\frac{1}{2^2}+\frac{1}{2^4}+\frac{1}{2^6}+....\left(< 0,2\right)\left(đcmp\right)\)

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

\(<\frac{1}{2^4}-\frac{1}{2^4}+\frac{1}{2^8}-\frac{1}{2^8}+...+\frac{1}{2^{4n}}-\frac{1}{2^{4n}}+...+\frac{1}{2^{2004}}-\frac{1}{2^{2004}}\)=0+0+0+...+0+....+0=0 <0,2

Vậy S<0,2

Ảo quá \(\frac{1}{4n-2}<\frac{1}{4n}\)

\(S=\left(\frac{1}{2^2}+...+\frac{1}{2^{2002}}\right)-\frac{1}{2^2}\left(\frac{1}{2^2}+...+\frac{1}{2^{2002}}\right)=\frac{3}{4}\left(\frac{1}{2^2}+...+\frac{1}{2^{2002}}\right)\)

\(S<0,2\Leftrightarrow\frac{3}{4}\left(\frac{1}{2^2}+...+\frac{1}{2^{2002}}\right)<0,2\Leftrightarrow\frac{1}{2^2}+...+\frac{1}{2^{2002}}<\frac{4}{15}\)

Ta có : \(2P-P=\frac{1}{2}+...+\frac{1}{2^{2001}}-\frac{1}{2^2}-...-\frac{1}{2^{2002}}=\frac{1}{2}-\frac{1}{2^{2002}}\) với \(P=\frac{1}{2^2}+...+\frac{1}{2^{2002}}\)

Thế mà P< 4/15 chịu

S = \(...\)

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

=> \(S.4+S=\left(1-\frac{1}{2^2}+\frac{1}{2^4}-...-\frac{1}{2^{2002}}\right)+\left(\frac{1}{2^2}-\frac{1}{2^4}+\frac{1}{2^6}-...-\frac{1}{2^{2004}}\right)\)

=> \(5S=1-\frac{1}{2^{2004}}<1\)

=> \(S<1:5=0,2\left(đpcm\right)\)

Vậy S < 0,2.

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

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

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

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

\(\Rightarrow\)5S<1 hay S<\(\dfrac{1}{5}\)=0,2(đpcm)

Mình nghĩ nó bé hơn \(0,2\) chứ nhỉ? Phạm Nguyễn Thục Anh