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\(\dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{2007^2}< \dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{2006.2007}\)
\(=\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{2006}-\dfrac{1}{2007}\)
\(=\dfrac{1}{4}-\dfrac{1}{2007}< \dfrac{1}{4}\)
\(\Rightarrow\dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{2007^2}< \dfrac{1}{4}\left(đpcm\right)\)
Vậy...
Ta có: \(A=\dfrac{1}{5^2}+\dfrac{2}{5^3}+...+\dfrac{11}{5^{12}}\)
\(\Rightarrow5A=\dfrac{1}{5}+\dfrac{2}{5^2}+...+\dfrac{11}{5^{11}}\)
\(\Rightarrow5A-A=\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^{11}}-\dfrac{11}{5^{12}}\)
\(\Rightarrow4A=\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^{11}}-\dfrac{11}{5^{12}}\)
\(\Rightarrow20A=1+\dfrac{1}{5}+...+\dfrac{1}{5^{10}}-\dfrac{11}{5^{11}}\)
\(\Rightarrow20A-4A=\left(1+\dfrac{1}{5}+...+\dfrac{1}{5^{10}}-\dfrac{11}{5^{11}}\right)-\left(\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^{11}}-\dfrac{11}{5^{12}}\right)\)
\(\Rightarrow16A=1-\dfrac{12}{5^{11}}+\dfrac{11}{5^{12}}< 1\)
\(\Rightarrow A< \dfrac{1}{16}\)
⇒5A=15+252+...+11511⇒5A=15+252+...+11511
⇒5A−A=15+152+...+1511−11512⇒5A−A=15+152+...+1511−11512
⇒4A=15+152+...+1511−11512⇒4A=15+152+...+1511−11512
⇒20A=1+15+...+1510−11511⇒20A=1+15+...+1510−11511
⇒20A−4A=(1+15+...+1510−11511)−(15+152+...+1511−11512)⇒20A−4A=(1+15+...+1510−11511)−(15+152+...+1511−11512)
⇒16A=1−12511+11512<1⇒16A=1−12511+11512<1
⇒A<116⇒A<116
Ta có : A=\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{9^2}\)
\(\Rightarrow A=\dfrac{1}{2.2}+\dfrac{1}{3.3}+\dfrac{1}{4.4}+...+\dfrac{1}{9.9}\)
\(\Rightarrow A=\dfrac{1}{2.2}+\dfrac{1}{3.3}+\dfrac{1}{4.4}+...+\dfrac{1}{9.9}\)<\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{8.9}\)
\(\Rightarrow A< \dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{8}-\dfrac{1}{9}\)
\(\Rightarrow A< \dfrac{1}{1}-\dfrac{1}{9}\)
\(\Rightarrow A< \dfrac{9}{9}-\dfrac{1}{9}\)
\(\Rightarrow A< \dfrac{8}{9}\) (1)
\(\Rightarrow A=\dfrac{1}{2.2}+\dfrac{1}{3.3}+\dfrac{1}{4.4}+...+\dfrac{1}{9.9}\)>\(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{9.10}\)
\(\Rightarrow A>\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{1}{10}\)
\(\Rightarrow A>\dfrac{1}{2}-\dfrac{1}{10}\)
\(\Rightarrow A>\dfrac{5}{10}-\dfrac{1}{10}\)
\(\Rightarrow A>\dfrac{4}{10}\)
\(\Rightarrow A>\dfrac{2}{5}\) (2)
Từ (1) và (2)\(\Rightarrow\dfrac{2}{5}< A< \dfrac{8}{9}\)
đó giúp mk đi mà
à, mk quên chưa nói là ai giúp mk sẽ được luôn 2SP đó
giúp mk nha
cảm ơn nhiều!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Kiyoko Vũ
a, xét từng đoạn 1 , 1/2 ,1/2^3 ,1/2^4 ,1/2^5 ,1/2^6
ta có
1 = 1
1/2 + 1/3 < 1/2 + 1/2 = 1
1/4 + 1/5 + .. + 1/7 < 1/4 +..+ 1/4 = 4/4 = 1
1/8 + 1/9 + .. + 1/15 < 1/8 + .. + 1/8 = 8/8 = 1
tương tự
1/16 +1/17 + .. + 1/31 < 1
1/32 + 1/33 + .. + 1/63 < 1
=> cộng lại => A < 6
b, Câu hỏi của trịnh quỳnh trang - Toán lớp 6 - Học toán với OnlineMath
Ta có: \(1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{19}-\dfrac{1}{20}\)
\(=1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{19}+\dfrac{1}{20}-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{20}\right)\)
\(=1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{19}+\dfrac{1}{20}-\left(1+\dfrac{1}{2}+...+\dfrac{1}{10}\right)\)
\(=1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{19}+\dfrac{1}{20}-1-\dfrac{1}{2}-...-\dfrac{1}{10}\)
\(=\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{20}\)
Vậy \(1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{19}-\dfrac{1}{20}=\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{20}\)
A=\(\dfrac{1}{5}+\dfrac{1}{5^2}+\dfrac{1}{5^3}+...+\dfrac{1}{5^{2014}}\)
5A=\(\dfrac{5}{5}+\dfrac{5}{5^2}+\dfrac{5}{5^3}+...+\dfrac{5}{5^{2014}}\)
5A=\(1+\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^{2013}}\)
5A-A=\(\left(1+\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^{2013}}\right)-\left(\dfrac{1}{5}+\dfrac{1}{5^2}+\dfrac{1}{5^3}+...+\dfrac{1}{5^{2014}}\right)\)4A=\(1-\dfrac{1}{5^{2014}}\)
4A=\(\dfrac{5^{2014}-1}{5^{2014}}\)
A=\(\dfrac{5^{2014}-1}{5^{2014}}:4\)
A=\(\dfrac{5^{2014}-1}{5^{2014}}.\dfrac{1}{4}\)
\(\Rightarrow\)A<\(\dfrac{1}{4}\)
Ta có:
A = \(\dfrac{1}{5}+\dfrac{1}{5^2}+\dfrac{1}{5^3}+....+\dfrac{1}{5^{2014}}\)
\(\Rightarrow\) 5A = 5\(\left(\dfrac{1}{5}+\dfrac{1}{5^2}+\dfrac{1}{5^3}+....+\dfrac{1}{5^{2014}}\right)\)
\(\Rightarrow\) 5A = \(\dfrac{5}{5}+\dfrac{5}{5^2}+\dfrac{5}{5^3}+....+\dfrac{5}{5^{2014}}\)
\(\Rightarrow\) 5A = \(1+\dfrac{1}{5}+\dfrac{1}{5^2}+....+\dfrac{1}{5^{2013}}\)
\(\Rightarrow\)\(\left(1+\dfrac{1}{5}+\dfrac{1}{5^2}+....+\dfrac{1}{5^{2013}}\right)\)-\(\left(\dfrac{1}{5}+\dfrac{1}{5^2}+\dfrac{1}{5^3}+....+\dfrac{1}{5^{2014}}\right)\) = 5A - A
\(\Rightarrow\)4A= 1 - \(\dfrac{1}{5^{2014}}\)
\(\Rightarrow\) A =\(\dfrac{5^{2014}-1}{5^{2014}}\) : 4
Vậy A =\(\dfrac{5^{2014}-1}{5^{2014}}\) : 4