Chứng minh rằng: \(\dfrac{1}{5}+\dfrac{1}{13}+\dfrac{1}{25}+...+\dfrac{1}{n^2+\left(n+1\right)^2}< \dfrac{1}{2}\)
giúp mk nha !! ^_^
Chứng minh rằng với mọi số tự nhiên n:
\(\dfrac{1}{5}+\dfrac{1}{13}+\dfrac{1}{25}...+\dfrac{1}{n^2+\left(n+1\right)^2}< \dfrac{9}{20}\)
\(a^2+\left(a+1\right)^2=a^2+a^2+2a+1\\ =2a^2+2a+1>2a\left(a+1\right)\\ \Rightarrow\dfrac{1}{a^2+\left(a+1\right)^2}< \dfrac{1}{2a\left(a+1\right)}\)
\(\dfrac{1}{5}+\dfrac{1}{13}+\dfrac{1}{25}+...+\dfrac{1}{n^2+\left(n+1\right)^{^2}}\\ =\dfrac{1}{1^2+2^2}+\dfrac{1}{2^2+3^2}+\dfrac{1}{3^2+4^2}+...+\dfrac{1}{n^2+\left(n+1\right)^2}\\ < \dfrac{1}{2.1.\left(1+2\right)}+\dfrac{1}{2.2\left(2+1\right)}+....+\dfrac{1}{2n\left(n+1\right)}\\ =\dfrac{1}{2}\left(\dfrac{1}{3}+\dfrac{1}{2.3}+...+\dfrac{1}{n\left(n+1\right)}\right)\\ =\dfrac{1}{2}\left(\dfrac{1}{3}+\dfrac{1}{2}-\dfrac{1}{n+1}\right)\\ =\dfrac{1}{2}\left(\dfrac{5}{6}-\dfrac{1}{n+1}\right)\\ =\dfrac{5}{12}-\dfrac{1}{2n+2}< \dfrac{5}{12}< \dfrac{9}{20}\)
Cho biểu thức \(A=\dfrac{1}{3}+\left(\dfrac{1}{3}\right)^2+\left(\dfrac{1}{3}\right)^3+...+\left(\dfrac{1}{3}\right)^{2019}+\left(\dfrac{1}{3}\right)^{2020}\). Chứng minh rằng A \(< \dfrac{1}{2}\)
Giúp mk đi, 23h là mk phải nộp rùi
\(A=\dfrac{1}{3}+\left(\dfrac{1}{3}\right)^2+\left(\dfrac{1}{3}\right)^3+...+\left(\dfrac{1}{3}\right)^{2019}+\left(\dfrac{1}{3}\right)^{2020}\)
\(\Rightarrow\dfrac{1}{3}A=\left(\dfrac{1}{3}\right)^2+\left(\dfrac{1}{3}\right)^3+...+\left(\dfrac{1}{3}\right)^{2021}\)
\(\Rightarrow\dfrac{2}{3}A=A-\dfrac{1}{3}A=\dfrac{1}{3}+\left(\dfrac{1}{3}\right)^2+\left(\dfrac{1}{3}\right)^3+...+\left(\dfrac{1}{3}\right)^{2020}-\left(\dfrac{1}{3}\right)^2-\left(\dfrac{1}{3}\right)^3-\left(\dfrac{1}{3}\right)^{2021}=\dfrac{1}{3}-\left(\dfrac{1}{3}\right)^{2021}< \dfrac{1}{3}\)
\(\Rightarrow A< \dfrac{1}{2}\)
chứng minh rằng \(S=\dfrac{1}{4^2}+\dfrac{1}{6^2}+\dfrac{1}{8^2}+...+\dfrac{1}{\left(2n\right)^2}< \dfrac{1}{4}\left(n\in N,n\ge2\right)\)
\(S=\dfrac{1}{2^2}\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{n^2}\right)\)
=>\(S< =\dfrac{1}{4}\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n-1}-\dfrac{1}{n}\right)\)
=>\(S< =\dfrac{1}{4}\cdot\left(1-\dfrac{1}{n}\right)=\dfrac{1}{4}\cdot\dfrac{n-1}{n}< =\dfrac{1}{4}\)
Cho A = \(\dfrac{\left(3\dfrac{2}{15}+\dfrac{1}{5}\right):2\dfrac{1}{2}}{\left(5\dfrac{3}{7}-2\dfrac{1}{4}\right):4\dfrac{43}{56}}\) ; B = \(\dfrac{1,2:\left(1\dfrac{1}{5}.1\dfrac{1}{4}\right)}{0,32+\dfrac{2}{25}}\)
Chứng minh rằng A= B
\(A=\dfrac{\left(3+\dfrac{2}{15}+\dfrac{1}{5}\right):\dfrac{5}{2}}{\left(5+\dfrac{3}{7}-2-\dfrac{1}{4}\right):\left(4+\dfrac{43}{56}\right)}\)
\(=\dfrac{\dfrac{10}{3}\cdot\dfrac{2}{5}}{\dfrac{89}{28}:\dfrac{267}{56}}=\dfrac{4}{3}:\dfrac{2}{3}=2\)
\(B=\dfrac{\dfrac{6}{5}:\left(\dfrac{6}{5}\cdot\dfrac{5}{4}\right)}{\dfrac{8}{25}+\dfrac{2}{25}}=\dfrac{\dfrac{6}{5}:\dfrac{3}{2}}{\dfrac{2}{5}}=2\)
Do đó: A=B
Chứng minh rằng :
a) \(\dfrac{1.3.5.....39}{21.22.23.....40}=\dfrac{1}{2^{20}}\)
b) \(\dfrac{1.3.5....\left(2n-1\right)}{\left(n+1\right)\left(n+2\right)\left(n+3\right)...2n}=\dfrac{1}{2^n}\) với \(n\in\) N*
a) Vế trái \(=\dfrac{1.3.5...39}{21.22.23...40}=\dfrac{1.3.5.7...21.23...39}{21.22.23....40}=\dfrac{1.3.5.7...19}{22.24.26...40}\)
\(=\dfrac{1.3.5.7....19}{2.11.2.12.2.13.2.14.2.15.2.16.2.17.2.18.2.19.2.20}\\ =\dfrac{1.3.5.7.9.....19}{\left(1.3.5.7.9...19\right).2^{20}}=\dfrac{1}{2^{20}}\left(đpcm\right)\)
b) Vế trái
\(=\dfrac{1.3.5...\left(2n-1\right)}{\left(n+1\right).\left(n+2\right).\left(n+3\right)...2n}\\ =\dfrac{1.2.3.4.5.6...\left(2n-1\right).2n}{2.4.6...2n.\left(n+1\right)\left(n+2\right)...2n}\\ =\dfrac{1.2.3.4...\left(2n-1\right).2n}{2^n.1.2.3.4...n.\left(n+1\right)\left(n+2\right)...2n}\\ =\dfrac{1}{2^n}.\\ \left(đpcm\right)\)
1) Chứng minh rằng: \(1+\dfrac{1}{2\sqrt{2}}+\dfrac{1}{3\sqrt{3}}+...+\dfrac{1}{n\sqrt{n}}< 2\sqrt{2}\left(n\in N\right)\)
2) Chứng minh rằng: \(\dfrac{2}{3}+\sqrt{n+1}< 1+\sqrt{2}+\sqrt{3}+...+\sqrt{n}< \dfrac{2}{3}\left(n+1\right)\sqrt{n}\)
3) \(2\sqrt{n}-3< \dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{n}}< 2\sqrt{n}-2\)
4) \(\dfrac{\sqrt{2}-\sqrt{1}}{2+1}+\dfrac{\sqrt{3}-\sqrt{2}}{3+2}+...+\dfrac{\sqrt{n+1}-\sqrt{n}}{n+1+n}< \dfrac{1}{2}\left(1-\dfrac{1}{\sqrt{n+1}}\right)\)
Chứng minh rằng : \(\dfrac{3}{4}+\dfrac{5}{36}+\dfrac{7}{114}+.....+\dfrac{2n+1}{n^2\left(n+1\right)^2}< 1\)
Cho A = \(\dfrac{1}{5}+\dfrac{1}{13}+\dfrac{1}{25}+.....+\dfrac{1}{n^2+\left(n+1\right)^2}\) . CMR A < \(\dfrac{1}{2}\)
Ta dễ dàng chứng minh được: \(n^2+\left(n+1\right)^2>2n\left(n+1\right)\)
Thật vậy:
\(n^2+\left(n+1\right)^2=n^2+n^2+2n+1=2n^2+2n+1>2n^2+2n=2n\left(n+1\right)\)Trở lại bài toán
\(A=\dfrac{1}{5}+\dfrac{1}{13}+\dfrac{1}{25}+...+\dfrac{1}{n^2+\left(n+1\right)^2}\)
\(A=\dfrac{1}{1^2+2^2}+\dfrac{1}{2^2+3^2}+\dfrac{1}{3^2+4^2}+....+\dfrac{1}{n^2+\left(n+1\right)^2}\)
\(A< \dfrac{1}{2.1.\left(1+1\right)}+\dfrac{1}{2.2.\left(2+1\right)}+\dfrac{1}{2.3.\left(3+1\right)}+....+\dfrac{1}{2n\left(n+1\right)}\)
\(A< \dfrac{1}{2.1.2}+\dfrac{1}{2.2.3}+\dfrac{1}{2.3.4}+....+\dfrac{1}{2n\left(n+1\right)}\)
\(A< \dfrac{1}{2}\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{n\left(n+1\right)}\right)\)
\(A< \dfrac{1}{2}\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+....+\dfrac{1}{n}-\dfrac{1}{n+1}\right)\)
\(A< \dfrac{1}{2}\left(1-\dfrac{1}{n+1}\right)\)
\(A< \dfrac{1}{2}-\dfrac{1}{2n+2}< \dfrac{1}{2}\left(đpcm\right)\)
chứng minh \(1^2+2^2+...+n^2=\dfrac{1}{3}n\left(n+\dfrac{1}{2}\right)\left(n+1\right)=\dfrac{1}{3}n^3+\dfrac{1}{2}n^2+\dfrac{1}{6}n\)
\(1^2+2^2+...+n^2=1+2\left(1+1\right)+...+n\left(n-1+1\right)=1+2+1.2+3+2.3+...+n+\left(n-1\right)n\)
\(=\left(1+2+3+...+n\right)+\left[1.2+2.3+...+\left(n-1\right)n\right]=\dfrac{\left(n+1\right)\left(\dfrac{n-1}{1}+1\right)}{2}+\dfrac{1.2.3+2.3.3+...+\left(n-1\right)n.3}{3}=\dfrac{n\left(n+1\right)}{2}+\dfrac{1.2.3+2.3.\left(4-1\right)+...+\left(n-1\right)n\left[\left(n+1\right)-\left(n-2\right)\right]}{3}\)
\(=\dfrac{n\left(n+1\right)}{2}+\dfrac{1.2.3-1.2.3+2.3.4-...-\left(n-2\right)\left(n-1\right)n+\left(n-1\right)n\left(n+1\right)}{3}\)
\(=\dfrac{n\left(n+1\right)}{2}+\dfrac{\left(n-1\right)n\left(n+1\right)}{3}=\dfrac{3n\left(n+1\right)+2\left(n-1\right)n\left(n+1\right)}{6}=\dfrac{2n^3+3n^2+n}{6}=\dfrac{1}{3}n^3+\dfrac{1}{2}n^2+\dfrac{1}{6}n=\dfrac{1}{3}n\left(n^2+\dfrac{3}{2}n+\dfrac{1}{2}\right)=\dfrac{1}{3}n\left(n+\dfrac{1}{2}\right)\left(n+1\right)\)