Cho \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2017^2}\)
\(B=\frac{2!}{3!}+\frac{2!}{4!}+\frac{2!}{5!}+...+\frac{2!}{2017!}\)
Chứng minh \(A+B< 2\)
K
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24 tháng 4 2017
a/ Ta có
\(200-\left(3+\frac{2}{3}+\frac{2}{4}+...+\frac{2}{100}\right)\)
\(=1+2\left(1-\frac{1}{3}\right)+2\left(1-\frac{1}{4}\right)+...+2\left(1-\frac{1}{100}\right)\)
\(=1+2\left(\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}\right)\)
\(=2\left(\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}\right)\)
Thế lại bài toán ta được:
\(\frac{200-\left(3+\frac{2}{3}+\frac{2}{4}+...+\frac{2}{100}\right)}{\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}}\)
\(=\frac{2\left(\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}\right)}{\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}}=2\)
24 tháng 4 2017
b/ Ta có:
A - B\(=\frac{-21}{10^{2016}}+\frac{12}{10^{2016}}+\frac{21}{10^{2017}}-\frac{12}{10^{2017}}\)
\(=\frac{9}{10^{2017}}-\frac{9}{10^{2016}}< 0\)
Vậy A < B
XO
26 tháng 10 2019
A = \(\frac{\frac{3}{4}-\frac{3}{11}+\frac{3}{13}}{\frac{5}{4}-\frac{5}{11}+\frac{5}{13}}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}}{\frac{5}{4}-\frac{5}{6}+\frac{5}{8}}\)
\(=\frac{3.\left(\frac{1}{4}-\frac{1}{11}+\frac{1}{13}\right)}{5.\left(\frac{1}{4}-\frac{1}{11}+\frac{1}{13}\right)}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}}{\frac{5}{2}.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}\right)}\)
\(=\frac{3}{5}+\frac{1}{\frac{5}{2}}\)
\(=\frac{3}{5}+\frac{2}{5}=1\)
XO
26 tháng 10 2019
b) B = \(\frac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6.8^4.3^5}-\frac{5^{10}.7^3:25^5.49}{\left(125.7\right)^3+5^9.14^3}\)
\(=\frac{2^{12}.3^5-\left(2^2\right)^6.\left(3^2\right)^2}{2^{12}.3^6+\left(2^3\right)^4.3^5}-\frac{5^{10}.7^3-\left(5^2\right)^5.7^2}{\left(5^3\right)^3.7^3+5^9.\left(7.2\right)^3}\)
\(=\frac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}-\frac{5^{10}.7^3-5^{10}-7^2}{5^9.7^3+5^9.7^3.2^3}\)
\(=\frac{2^{12}.3^4.\left(3-1\right)}{2^{12}.3^5\left(3+1\right)}-\frac{5^{10}.7^2.\left(7-1\right)}{5^9.7^3\left(1+2^3\right)}\)
\(=\frac{1}{3.2}-\frac{5.2}{7.3}\)
\(=\frac{7}{3.2.7}-\frac{5.2.2}{7.3.2}\)
\(=\frac{7}{42}-\frac{20}{42}\)
\(=-\frac{13}{42}\)
7 tháng 4 2019
\(C=\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\)
\(>\frac{a}{a+b+c}+\frac{b}{a+b+c}+\frac{c}{a+b+c}=1\)
\(D< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2016.2017}\)
\(\Rightarrow D< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2016}-\frac{1}{2017}\)
\(\Rightarrow D< 1-\frac{1}{2017}< 1\)
Vậy C > D
CX
13 tháng 3 2020
Ta có : \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2018^2}< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2017.2018}\)
Xét B = \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2017.2018}\)
= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}\)
=\(1-\frac{1}{2018}\)
Xét : \(\frac{2018}{2018}=1\)=) B < 1
khoan hình như sai đề
A=1/2^2 + 1/3^2 + 1/4^2 + ... + 1/2017^2
A < 1/1.2 + 1/2.3 + 1/3.4 + ... + 1/2016.2017
A < 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/2016 - 1/2017
A < 1 - 1/2017 < 1 (1)
B = 2!/3! + 2!/4! + 2!/5! + ... + 2!/2017!
B = 2!.(1/3! + 1/4! + 1/5! + ... + 1/2017!)
B < 2.(1/2.3 + 1/3.4 + 1/4.5 + ... + 1/2016.2017)
B < 2.(1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + ... + 1/2016 - 1/2017)
B < 2.(1/2 - 1/2017) < 2.1/2 = 1 (2)
Từ (1) và (2) => A + B < 2 (đpcm)