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Đặt A = \(1+\frac{1}{1.2}+\frac{1}{1.2.3}+\frac{1}{1.2.3.4}+...+\frac{1}{1.2.3....n}\)
Ta có: \(\frac{1}{1.2}=\frac{1}{1.2}\)
\(\frac{1}{1.2.3}=\frac{1}{2.3}\)
\(\frac{1}{1.2.3.4}< \frac{1}{3.4}\)
..............
\(\frac{1}{1.2.3....n}< \frac{1}{\left(n-1\right)n}\)
Cộng vế với vế ta được:
\(A< 1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right)n}=1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n-1}-\frac{1}{n}=1+1-\frac{1}{n}=2-\frac{1}{n}< 2\)(đpcm)
A\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}-2\cdot\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)\)
\(=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)
Ta thấy
A\(=\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{75}\right)+\left(\frac{1}{76}+\frac{1}{77}+...+\frac{1}{100}\right)\)
=> A> \(\frac{1}{75}\cdot25+\frac{1}{100}\cdot25\)
=>A > 7/12
A\(=\frac{1}{51}+...+\frac{1}{60}+\left(\frac{1}{61}+...+\frac{1}{70}\right)+\left(\frac{1}{71}+...+\frac{1}{80}\right)+\left(\frac{1}{81}+...+\frac{1}{90}\right)+\left(\frac{1}{91}+...+\frac{1}{100}\right)\)>\(\frac{1}{60}\cdot10+\frac{1}{70}\cdot10+\frac{1}{80}\cdot10+\frac{1}{90}\cdot10+\frac{1}{100}\cdot10\)
>\(\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}\)
>1/6 *5
>5/6(chac la chuan roi day)
ta có:
1.2-1/2!+2.3-1/3!+3.4-1/4!+...+99.100-1/100!
=1.2/2!-1/2!+2.3/3!-13!+...+99.100-1/100!
=(1.2/2!+2.3/3!+3.4-4!+...+99.100/100!)-(1/2!+1/3!+...+1/100!)
=(1+1+1/2+...+1/98!)_(1/2!+1/3!+...+1/100!)
=2-1/99!-1/100!<2
Ta xét :
\(\frac{1.2-1}{2!}+\frac{2.3-1}{3!}+\frac{3.4-1}{4!}+...+\frac{99.100-1}{100!}\)
\(=\frac{1.2}{2!}-\frac{1}{2!}+\frac{2.3}{3!}-\frac{1}{3!}+...+\frac{99.100}{100!}-\frac{1}{100!}\)
\(=\left(\frac{1.2}{2!}+\frac{2.3}{3!}+\frac{3.4}{4!}...+\frac{99.100}{100!}\right)-\left(\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+...+\frac{1}{100!}\right)\)
\(=1+1-\frac{1}{99}-\frac{1}{100}\)
\(=2-\frac{1}{99}-\frac{1}{100}< 2\)
\(\RightarrowĐPCM\)
3n+2 - 2n+2 +3n - 2n = 3n . 32 - 2n. 22 +3n -2n
= 3n(32+1) - (2n.22 +2n)
=3n . 10 - 2n .5
=3n.10 - 2n-1 .2 .5
= 3n.10 - 2n-1 .10
= 10(3n - 2n-1)
vì 10 chia hết cho 10 nên 10(3n-2n-1) chia hết cho 10
=> 3n+2 - 2n+2 +3n -2n chia hết cho 10
Ai làm nhanh nhất mình sẽ **** xin cảm ơn các bạn mình đang cần gấp
a. \(\frac{1}{2}\) - ( \(\frac{1}{3}\) + \(\frac{1}{4}\) ) < x < \(\frac{1}{48}\) - ( \(\frac{1}{16}\) - \(\frac{1}{6}\) )
\(\frac{1}{2}\) - \(\frac{7}{12}\) < x < \(\frac{1}{48}\) - \(\frac{-5}{48}\)
\(\frac{-1}{12}\) < x < \(\frac{1}{8}\)
Đề bài yêu cầu tìm x thuộc tập hợp gì bạn ơi. Bạn viết thiếu rồi .
Ta có:
\(\frac{1}{1.2.3.4}<\frac{1}{3.4}\)
\(\frac{1}{1.2.3.4.5}<\frac{1}{4.5}\)
\(...\)
\(\frac{1}{1.2.3...n}<\frac{1}{\left(n-1\right)n}\)
\(\Rightarrow1+\frac{1}{1.2}+\frac{1}{1.2.3}+\frac{1}{1.2.3.4}+....+\frac{1}{1.2.3...n}<1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{\left(n-1\right)n}\)
\(\Rightarrow1+\frac{1}{1.2}+\frac{1}{1.2.3}+\frac{1}{1.2.3.4}+....+\frac{1}{1.2.3...n}<1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n-1}-\frac{1}{n}\)
\(\Rightarrow1+\frac{1}{1.2}+\frac{1}{1.2.3}+\frac{1}{1.2.3.4}+....+\frac{1}{1.2.3...n}<1+\frac{1}{n}-\frac{1}{n-1}\)
\(\Rightarrow1+\frac{1}{1.2}+\frac{1}{1.2.3}+\frac{1}{1.2.3.4}+....+\frac{1}{1.2.3...n}<1+\frac{n-1}{n}\)
Vì \(\frac{n-1}{n}<1\Rightarrow\frac{n-1}{n}+1<2\)
\(\Rightarrow1+\frac{1}{1.2}+\frac{1}{1.2.3}+\frac{1}{1.2.3.4}+....+\frac{1}{1.2.3...n}<2\)