Chứng minh rằng: \(\frac{1}{3}\)+\(\frac{1}{4}\)+\(\frac{1}{5}\)+...+\(\frac{1}{130}\)>3
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Chứng minh rằng:\(A=\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{130}\)lớn hơn 3
Cac thua so trong A deu nho hon hoac bang 1/130
=>(1/3)+(1/3)+(1/3)+(1/3)+(1/3)+(1/3)+(1/3)+(1/130)+(1/130)+...+(1/130)(121 p/s 1/130)<A
7*(1/3)+121*(1/130)<A
647/195<A
3*(62/195)<A
=>A>3
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24 tháng 2 2017 lúc 16:22
dạng 1 : so sánh a) P frac{1}{1^2}+frac{1}{2^2}+frac{1}{3^2}+...+frac{1}{2013^2}+frac{1}{2014^2}và Q 1frac{3}{4}dạng 2 : toán chứng minh 1. cho S frac{1}{101}+frac{1}{102}+...+frac{1}{130}chứng minh rằng : frac{1}{4} S frac{91}{330}2. cho S frac{5}{20}+frac{5}{21}+frac{5}{22}+...+frac{5}{49}. CMR : 3 S 83. CMR : 1+frac{1}{2}+frac{1}{3}+...+frac{1}{2^{1999}}1000
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dạng 1 : so sánh
a) P = \(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2013^2}+\frac{1}{2014^2}\)và Q = \(1\frac{3}{4}\)
dạng 2 : toán chứng minh
1. cho S = \(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{130}\)chứng minh rằng : \(\frac{1}{4}< S< \frac{91}{330}\)
2. cho S = \(\frac{5}{20}+\frac{5}{21}+\frac{5}{22}+...+\frac{5}{49}\). CMR : 3 < S < 8
3. CMR : \(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2^{1999}}>1000\)
2.a) Vào question 126036
b) Vào question 68660
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Chứng minh rằng :
\(100-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)=\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+\frac{4}{5}+...+\frac{99}{100}\)
Ta có :\(100-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)
=\(\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}=\)\(\left(1-1\right)+\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{3}\right)\)\(+...+\left(1-\frac{1}{100}\right)\)
=\(\left(1+1+1+....+1\right)\)\(-\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\)
= \(99-\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\)
= \(100-1-\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\)
=\(100-\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\)= vế trên (đpcm)
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\(S=100-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)
\(S=\left(1+1+...+1\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)
\(S=\left(1-1\right)+\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{3}\right)+...+\left(1-\frac{1}{100}\right)\)
\(S=\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}\)
\(\RightarrowĐPCM\)
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\(\text{Chứng minh rằng:}\)
\(\frac{1}{2^3}+\frac{1}{3^3}+\frac{1}{4^3}+\frac{1}{5^3}+...+\frac{1}{n^3}< \frac{1}{4}\)
Chứng minh rằng \(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\frac{4}{5!}+....+\frac{99}{100!}< 1\)1
Ta có : \(VT=\frac{2-1}{2!}+\frac{3-1}{3!}+\frac{4-1}{4!}+...+\frac{100-1}{100!}\)
\(=1-\frac{1}{2!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{3!}-\frac{1}{4!}+...+\frac{1}{99!}-\frac{1}{100!}\)
\(=1-\frac{1}{100!}< 1\)
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\(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\frac{4}{5!}+...+\frac{99}{100!}=\frac{2-1}{2!}+\frac{3-1}{3!}+\frac{4-1}{4!}+\frac{5-1}{5!}+...+\frac{100-1}{100!}\)
\(=\frac{2}{1.2}-\frac{1}{2!}+\frac{3}{1.2.3}-\frac{1}{3!}+\frac{4}{1.2.3.4}-\frac{1}{4!}+\frac{5}{1.2.3.4.5}-\frac{1}{5!}+...+\frac{100}{1.2...99.100}-\frac{1}{100!}\)
\(=\frac{1}{1}-\frac{1}{2!}+\frac{1}{1.2}-\frac{1}{3!}+\frac{1}{1.2.3}-\frac{1}{4!}+\frac{1}{1.2.3.4}-\frac{1}{5!}+...+\frac{1}{1.2...99}-\frac{1}{100!}\)
\(=1-\frac{1}{2!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{3!}-\frac{1}{4!}+\frac{1}{4!}-\frac{1}{5!}+...+\frac{1}{99!}-\frac{1}{100!}\)
\(=1-\frac{1}{100!}< 1\)
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Ta có : 1 = 2 - 1
2 = 3 - 1
...
99 = 100 - 1
Gọi tổng kia là S
Ta có : 2-1/2! + 3-1/3! + 4-1/4! + ... + 100-1/100!
S = 2/2! - 1/2 ! + 3/3 ! - 1/3! + 4/4! - 1/4! + ... + 100/100! - 1/100!
S = ......
S = 1 - 1/100! < 1
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Bài 1:a, Cho Sfrac{1}{2^2}+frac{1}{3^2}+frac{1}{4^2}+...+frac{1}{9^2} .Chứng minh rằng frac{2}{5} S frac{8}{9}b, Tìm x thuộc z để phân số frac{x^2-5x-1}{x+2}có giá trị là số nguyênc, Chứng minh rằng left(frac{7}{65}+1right)left(frac{7}{84}+1right)left(frac{7}{105}+1right)left(frac{7}{124}+1right)...left(frac{7}{153+1}right)left(frac{7}{560}+1right) 2d, Chứng minh rằng frac{1}{3}-frac{2}{3^2}+frac{3}{3^3}-frac{4}{3^4}+frac{5}{3^5}-...+frac{99}{3^{99}}-frac{100}{3^{100}} frac{3}{16}
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Bài 1:
a, Cho S=\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}\) .Chứng minh rằng \(\frac{2}{5}< S< \frac{8}{9}\)
b, Tìm x thuộc z để phân số \(\frac{x^2-5x-1}{x+2}\)có giá trị là số nguyên
c, Chứng minh rằng \(\left(\frac{7}{65}+1\right)\left(\frac{7}{84}+1\right)\left(\frac{7}{105}+1\right)\left(\frac{7}{124}+1\right)...\left(\frac{7}{153+1}\right)\left(\frac{7}{560}+1\right)< 2\)
d, Chứng minh rằng \(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+\frac{5}{3^5}-...+\frac{99}{3^{99}}-\frac{100}{3^{100}}< \frac{3}{16}\)
1.Chứng minh rằng: \(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^3.4^2}+...+\frac{19}{9^2.10^2}< 1\)
2.Chứng minh rằng: \(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+...+\frac{100}{3^{100}}< \frac{3}{4}\)
Làm nhanh giúp mình nhé mọi người !!!
Bài 1:
Ta có:
\(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}\)
\(=\frac{3}{1.4}+\frac{5}{4.9}+\frac{7}{9.16}+...+\frac{19}{81.100}\)
\(=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+...+\frac{1}{81}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
Mà \(\frac{99}{100}< 1\)
\(\Rightarrow\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}< 1\left(đpcm\right)\)
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Có phải ở sách NCPT ko bn
Bài 2: Đặt \(B=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+...+\frac{100}{3^{100}}\)
\(3B=1+\frac{2}{3}+\frac{3}{3^2}+\frac{4}{3^3}+...+\frac{100}{3^{99}}\)
\(3B-B=\left(1+\frac{2}{3}+\frac{3}{3^2}+\frac{4}{3^3}+...+\frac{100}{3^{99}}\right)-\left(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+...+\frac{100}{3^{100}}\right)\)
\(2B=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
\(6B=3+1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}-\frac{100}{3^{99}}\)
\(6B-2B=\left(3+1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}-\frac{100}{3^{99}}\right)-\left(1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\right)\)
\(4B=3-\frac{100}{3^{99}}-\frac{1}{3^{99}}+\frac{100}{3^{100}}\)
\(4B=3-\frac{300}{3^{100}}-\frac{3}{3^{100}}+\frac{100}{3^{100}}\)
\(4B=3-\frac{303}{3^{100}}+\frac{100}{3^{100}}\)
\(4B=3-\frac{203}{3^{100}}< 3\)
\(B< \frac{3}{4}\left(đpcm\right)\)
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bài 1: tính A:=\(\frac{1}{2}-\frac{2}{3}+\frac{3}{4}-\frac{4}{5}+\frac{5}{6}-\frac{6}{7}-\frac{5}{6}+\frac{4}{5}-\frac{3}{4}+\frac{2}{3}-\frac{2}{3}-\frac{1}{2}\)
Bài 2: Cho B=\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+.....+\frac{1}{49}-\frac{1}{50}\)
Chứng minh rằng: \(\frac{7}{12}< A< \frac{5}{6}\)
1. Chứng Minh Rằng \(\frac{1}{3^1}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+.....+\frac{100}{3^{100}}<\frac{3}{4}\)
2. Chứng Minh Rằng \(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{2015.2016}=\frac{1}{1009}+\frac{1}{1010}+\frac{1}{1011}+...+\frac{1}{2012}\)
2.
\(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{2015.2016}\)
\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{2015}-\frac{1}{2016}\)
\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{2015}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2016}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2016}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1008}\right)\)
\(=\frac{1}{1009}+\frac{1}{1010}+\frac{1}{1011}+...+\frac{1}{2016}\)
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