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Đề bài đúng, trích trong đề thi HSG nên đừng nói là sai đề nha!!!
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\(1)C=\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+...+\dfrac{1}{162}\)
\(3C=1+\dfrac{1}{3}+\dfrac{1}{9}+...+\dfrac{1}{54}\)
\(3C-C=\left(1+\dfrac{1}{3}+\dfrac{1}{9}+...+\dfrac{1}{54}\right)-\left(\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+...+\dfrac{1}{162}\right)\)
\(2C=1-\dfrac{1}{162}\)
\(2C=\dfrac{161}{162}\)
\(C=\dfrac{161}{162}.\dfrac{1}{2}\)
\(C=\dfrac{161}{324}\)
\(2)A=\dfrac{1}{2}+\dfrac{1}{8}+\dfrac{1}{32}+\dfrac{1}{128}+\dfrac{1}{512}\)
\(2A=1+\dfrac{1}{2}+\dfrac{1}{8}+\dfrac{1}{32}+\dfrac{1}{128}\)
\(2A-A=\left(1+\dfrac{1}{2}+\dfrac{1}{8}+\dfrac{1}{32}+\dfrac{1}{128}\right)-\left(\dfrac{1}{2}+\dfrac{1}{8}+\dfrac{1}{32}+\dfrac{1}{128}+\dfrac{1}{512}\right)\)
\(A=1-\dfrac{1}{512}=\dfrac{511}{512}\)
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Giải
\(\frac{1}{b}-\frac{1}{b+1}=\frac{b+1-b}{b\left(b+1\right)}=\frac{1}{b\left(b+1\right)}< \frac{1}{b.b}=\frac{1}{b^2}\)
Vậy \(\frac{1}{b^2}>\frac{1}{b}-\frac{1}{b+1}\) ( 1 )
\(\frac{1}{b-1}-\frac{1}{b}=\frac{b-b+1}{b\left(b-1\right)}=\frac{1}{b\left(b-1\right)}>\frac{1}{b.b}=\frac{1}{b^2}\)
Vậy \(\frac{1}{b^2}< \frac{1}{b-1}-\frac{1}{b}\) ( 2 )
Từ ( 1 ) và ( 2 ) suy ra \(\frac{1}{b}-\frac{1}{b+1}< \frac{1}{b^2}< \frac{1}{b-1}-\frac{1}{b}\left(đpcm\right)\)