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Ta có :
\(D=\dfrac{1}{3}+\dfrac{2}{3^2}+\dfrac{3}{3^3}+......................+\dfrac{100}{3^{100}}+\dfrac{101}{3^{101}}\)
\(3D=1+\dfrac{2}{3}+\dfrac{3}{3^2}+.....................+\dfrac{100}{3^{99}}\)
\(3D-D=\left(1+\dfrac{2}{3}+\dfrac{3}{3^2}+...................+\dfrac{101}{3^{101}}\right)-\left(\dfrac{1}{3}+\dfrac{2}{3^2}+..............+\dfrac{100}{3^{99}}\right)\)\(2D=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...............+\dfrac{1}{3^{99}}-\dfrac{100}{3^{100}}\)
\(6D=3+1+\dfrac{1}{3}+................+\dfrac{1}{3^{98}}-\dfrac{100}{3^{99}}\)
\(6D-2D=\left(3+1+\dfrac{1}{3}+.............+\dfrac{1}{3^{99}}-\dfrac{100}{3^{100}}\right)-\left(1+\dfrac{1}{3}+..........+\dfrac{1}{3^{99}}-\dfrac{100}{3^{99}}\right)\)\(4D=3-\dfrac{100}{3^{99}}-\dfrac{1}{3^{99}}+\dfrac{100}{3^{100}}\)
\(4D=3-\dfrac{300}{3^{100}}-\dfrac{3}{3^{100}}+\dfrac{100}{3^{100}}\)
\(4D=3-\dfrac{203}{3^{100}}< 3\)
\(\Rightarrow D< \dfrac{3}{4}\rightarrowđpcm\)
~ Chúc bn học tốt ~
cau 1
de a dat gia tri lon nhat suy ra5a-17/4a-23 lon nhat
suy ra 4a-23 phai nho nhat khac 0 va la so nguyen duong
suy ra 4a-23=1
suy ra 4a=1+23=24
suy ra a=24 chia 4=6
vay de a nho nhat thi a=6
Ta có: \(K=\dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{99^2}+\dfrac{1}{100^2}< \dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{99.100}\)
\(=\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{100}=\dfrac{1}{3}-\dfrac{1}{100}< \dfrac{1}{3}\) (1)
\(K=\dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{100^2}>\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{100.101}\)
\(=\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{100}-\dfrac{1}{101}=\dfrac{1}{4}-\dfrac{1}{101}>\dfrac{1}{5}\) (2)
Từ (1), (2) \(\Rightarrow\dfrac{1}{5}< K< \dfrac{1}{3}\left(đpcm\right)\)
\(K=\dfrac{9-5}{3}+\dfrac{2.9-5}{3^2}+\dfrac{3.9-5}{3^3}+...+\dfrac{101.9-5}{3^{101}}\)
\(K=\dfrac{9}{3}+\dfrac{2.9}{3^2}+\dfrac{3.9}{3^3}+...+\dfrac{101.9}{3^{101}}-5\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{101}}\right)\)
\(K=9\left(\dfrac{1}{3}+\dfrac{2}{3^2}+\dfrac{3}{3^3}+...+\dfrac{101}{3^{101}}\right)-5\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{101}}\right)\)
\(K=9A-5B\)
Xét \(A=\dfrac{1}{3}+\dfrac{2}{3^2}+\dfrac{3}{3^3}+...+\dfrac{101}{3^{101}}\) (1)
\(\Rightarrow\dfrac{1}{3}A=\dfrac{1}{3^2}+\dfrac{2}{3^3}+\dfrac{3}{3^4}+...+\dfrac{100}{3^{101}}+\dfrac{101}{3^{102}}\) (2)
Trừ vế với vế (1) cho (2):
\(\dfrac{2}{3}A=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{101}}-\dfrac{101}{3^{102}}=B-\dfrac{101}{3^{102}}\)
\(\Rightarrow A=\dfrac{3}{2}\left(B-\dfrac{101}{3^{102}}\right)\Rightarrow K=\dfrac{27}{2}\left(B-\dfrac{101}{3^{102}}\right)-5B\)
\(\Rightarrow K=\dfrac{17}{2}B-\dfrac{27}{2}.\dfrac{101}{3^{102}}\)
Xét \(B=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{101}}\)
\(\Rightarrow3B=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{90}}+\dfrac{1}{3^{100}}\)
\(\Rightarrow3B-1+\dfrac{1}{3^{101}}=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{101}}=B\)
\(\Rightarrow2B=1-\dfrac{1}{3^{101}}\Rightarrow B=\dfrac{1}{2}-\dfrac{1}{2}.\dfrac{1}{3^{101}}\)
\(\Rightarrow K=\dfrac{17}{2}\left(\dfrac{1}{2}-\dfrac{1}{2}.\dfrac{1}{3^{101}}\right)-\dfrac{27}{2}.\dfrac{101}{3^{102}}\)
\(\Rightarrow K=\dfrac{17}{4}-\dfrac{1}{3^{101}}\left(\dfrac{17}{4}+\dfrac{27.101}{6}\right)< \dfrac{17}{4}\) (đpcm)
good job