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1) Tính C
\(C=\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+....+\frac{n-1}{n!}\)
\(=\frac{2-1}{2!}+\frac{3-1}{3!}+\frac{4-1}{4!}+...+\frac{n-1}{n!}\)
\(=1-\frac{1}{2!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{3!}-\frac{1}{4!}+...+\frac{1}{\left(n-1\right)!}-\frac{1}{n!}\)
\(=1-\frac{1}{n!}\)
3) a) Ta có : \(P=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{199}-\frac{1}{200}\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{199}+\frac{1}{200}-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{200}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{199}+\frac{1}{200}-1-\frac{1}{2}-\frac{1}{3}-...-\frac{1}{100}\)
\(=\frac{1}{101}+\frac{1}{102}+....+\frac{1}{199}+\frac{1}{200}\left(đpcm\right)\)
Ta luôn chứng minh được: Nếu \(\frac{a}{b}>1\Leftrightarrow\frac{a}{b}>\frac{a+1}{b+1}\)và \(\frac{a}{b}< \frac{a-1}{b-1}\)
Áp dụng điều trên ta có:
\(S=\frac{2}{1}.\frac{4}{3}.\frac{6}{5}...\frac{200}{199}\)
=> \(S>\frac{3}{2}.\frac{5}{4}.\frac{7}{6}...\frac{201}{200}\)
=> \(S^2>\frac{2}{1}.\frac{3}{2}.\frac{4}{3}.\frac{5}{4}.\frac{6}{5}.\frac{7}{6}...\frac{200}{199}.\frac{201}{200}\)
=> S2 > 201 > 200 (1)
\(S=\frac{2}{1}.\frac{4}{3}.\frac{6}{5}...\frac{200}{199}\)
=> \(S< \frac{2}{1}.\frac{3}{2}.\frac{5}{4}...\frac{199}{198}\)
=> \(S^2< \frac{2}{1}.\frac{2}{1}.\frac{3}{2}.\frac{4}{3}.\frac{5}{4}.\frac{6}{5}...\frac{199}{198}.\frac{200}{199}\)
=> \(S^2< 400\)(2)
Từ (1) và (2) => 200 < S2 < 400 (đpcm)
a, \(A=\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)...\left(\frac{1}{200}-1\right)\)
\(-A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)...\left(1-\frac{1}{200}\right)\)
\(-A=\frac{1}{2}\cdot\frac{2}{3}\cdot...\cdot\frac{199}{200}\)
\(-A=\frac{1}{200}\)
\(A=\frac{-1}{200}>\frac{-1}{199}\)