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Bài làm :
Ta có :
\(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+...+\frac{n}{\left(n+1\right)!}\)
\(=\frac{1}{1.2}+\frac{2}{1.2.3}+\frac{3}{1.2.3.4}+...+\frac{n}{1.2.3...\left(n+1\right)}\)
\(=\frac{2-1}{1.2}+\frac{3-1}{1.2.3}+\frac{4-1}{1.2.3.4}+...+\frac{n+1-1}{1.2.3...\left(n+1\right)}\)
\(=1-\frac{1}{1.2}+\frac{1}{1.2}-\frac{1}{1.2.3}+\frac{1}{1.2.3}-\frac{1}{1.2.3.4}+...+\frac{1}{1.2.3.4..n}-\frac{1}{1.2.3.4...\left(n+1\right)}\)
\(=1-\frac{1}{1.2.3.4...\left(n+1\right)}\)
\(\text{Vì : }\frac{1}{1.2.3.4...\left(n+1\right)}>0\Rightarrow1-\frac{1}{1.2.3.4...\left(n+1\right)}< 1\)
=> Điều phải chứng minh
Ta có : \(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+...+\frac{n}{\left(n+1\right)!}=\frac{1}{1.2}+\frac{2}{1.2.3}+\frac{3}{1.2.3.4}+...+\frac{n}{1.2.3...\left(n+1\right)}\)
\(=\frac{2-1}{1.2}+\frac{3-1}{1.2.3}+\frac{4-1}{1.2.3.4}+...+\frac{n+1-1}{1.2.3....\left(n+1\right)}\)
\(=1-\frac{1}{1.2}+\frac{1}{1.2}-\frac{1}{1.2.3}+\frac{1}{1.2.3}-\frac{1}{1.2.3.4}+...+\frac{1}{1.2.3.4..n}-\frac{1}{1.2.3.4...\left(n+1\right)}\)
\(=1-\frac{1}{1.2.3.4...\left(n+1\right)}< 1\left(\text{đpcm}\right)\)
Ta có :
\(A=\frac{1.2-1}{2!}+\frac{2.3-1}{3!}+...+\frac{\left(n-1\right)n-1}{n!}\)
\(=\frac{1.2}{2!}-\frac{1}{2!}+\frac{2.3}{3!}-\frac{1}{3!}+\frac{3.4}{4!}-\frac{1}{4!}+...+\frac{\left(n-1\right)n}{n!}-\frac{1}{n!}\)
\(=1-\frac{1}{2!}+1-\frac{1}{3!}+\frac{1}{2!}-\frac{1}{4}!+\frac{1}{3!}-\frac{1}{5!}+\frac{1}{4!}-...+\frac{1}{\left(n-2\right)!}-\frac{1}{n!}\)
\(=2-\frac{1}{n!}< 2\)
Vậy ...
AFK
đm đề kiểu gì đấy n là STN và n>1 thì \(\frac{1}{n!}>0\)nhé
1. D= 1/3 + 1/3.4 + 1/3.4.5 + 1/3.4.5....n < 1/2 + 1/3.4 + 1/4.5 + ...+ 1/ n.(n-1)
=> còn lại thì bạn có thể tự chứng minh
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 : \(P=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{199}-\frac{1}{200}\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{199}+\frac{1}{200}-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{200}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{199}+\frac{1}{200}-1-\frac{1}{2}-\frac{1}{3}-...-\frac{1}{100}\)
\(=\frac{1}{101}+\frac{1}{102}+....+\frac{1}{199}+\frac{1}{200}\left(đpcm\right)\)