Chứng minh\(B=1-\dfrac{1}{2^2}-\dfrac{1}{3^2}-\dfrac{1}{4^2}-...-\dfrac{1}{2004^2}>\dfrac{1}{2004}\)
Chứng minh \(S=\dfrac{1}{2^2}-\dfrac{1}{2^4}+\dfrac{1}{2^6}-...+\dfrac{1}{2^{4n-2}}-\dfrac{1}{2^{4n}}+...+\dfrac{1}{2^{2002}}-\dfrac{1}{2^{2004 }}< 0.2\)
\(\dfrac{1}{2^2}-\dfrac{1}{2^4}+\dfrac{1}{2^6}-...+\dfrac{1}{2^{4n-2}}-\dfrac{1}{2^{4n}}+...+\dfrac{1}{2^{2002}}-\dfrac{1}{2^{2004}}\)<0,2
Tính tổng đại số
\(A=\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{2}{3}+\dfrac{1}{4}+\dfrac{2}{4}+\dfrac{3}{4}-\dfrac{1}{5}-\dfrac{2}{5}-\dfrac{3}{5}-\dfrac{4}{5}+...+\dfrac{1}{10}+\dfrac{2}{10}+...+\dfrac{9}{10}\)
\(B=\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{2}{3}+\dfrac{1}{4}+\dfrac{2}{4}+\dfrac{3}{4}+...+\dfrac{1}{n}+\dfrac{2}{n}+...+\dfrac{n-1}{n}\)\(\left(n\in Z,n\ge2\right)\)
Cho M = \(1-\dfrac{1}{2}-\dfrac{1}{2^2}-\dfrac{1}{2^3}-\dfrac{1}{2^4}-....-\dfrac{1}{2^{10}}\) . So sánh M với \(\dfrac{1}{2^{11}}\)
B = \(\dfrac{1}{2}-\dfrac{1}{2^2}+\dfrac{1}{2^3}-\dfrac{1}{2^4}+..........+\dfrac{1}{2^99}-\dfrac{1}{2^100}\)
Tính :
1) C = \(\left(\dfrac{1}{200^2}-1\right)\left(\dfrac{1}{199^2}-1\right)...\left(\dfrac{1}{101^2}-1\right)\)
2) \(D=\dfrac{1}{1-\dfrac{1}{1-2^{-1}}}+\dfrac{1}{1+\dfrac{1}{1+2^{-1}}}\)
cho A=(\(\dfrac{1}{2^2}-1\))(\(\dfrac{1}{3^2}-1\))(\(\dfrac{1}{2^2}-1\))...........(\(\dfrac{1}{100^2}-1\)).SO sánh A với \(\dfrac{-1}{2}\)
Chứng minh:
\(A=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{18.19.20}< \dfrac{1}{4}\)
\(B=\dfrac{36}{1.3.5}+\dfrac{36}{5.7.9}+\dfrac{36}{9.11.13}+...+\dfrac{36}{25.27.29}< 3\)
\(C=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{n^2}\in< 1\left(n\in N,n\ge2\right)\)
\(D=\dfrac{1}{4^2}+\dfrac{1}{6^2}+\dfrac{1}{8^2}+...+\dfrac{1}{\left(2n\right)^2}< 4\left(n\in N,n\ge2\right)\)
\(E=\dfrac{2!}{3!}+\dfrac{2!}{4!}+\dfrac{2!}{5!}+...+\dfrac{2!}{n!}< 1\left(n\in N,n\ge3\right)\)
A=\(\dfrac{1}{1.2^2}+\dfrac{1}{2.3^2}+\dfrac{1}{3.4^2}+...+\dfrac{1}{49.50^2}\)
B=\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{50^2}\)
CM A<B
Tính\(B=\dfrac{1}{2}-\dfrac{1}{2^2}+\dfrac{1}{2^3}-\dfrac{1}{2^4}+.........+\dfrac{1}{2^{99}}-\dfrac{1}{2^{100}}\)