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\(K=\dfrac{9-5}{3}+\dfrac{2.9-5}{3^2}+\dfrac{3.9-5}{3^3}+...+\dfrac{101.9-5}{3^{101}}\)
\(K=\dfrac{9}{3}+\dfrac{2.9}{3^2}+\dfrac{3.9}{3^3}+...+\dfrac{101.9}{3^{101}}-5\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{101}}\right)\)
\(K=9\left(\dfrac{1}{3}+\dfrac{2}{3^2}+\dfrac{3}{3^3}+...+\dfrac{101}{3^{101}}\right)-5\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{101}}\right)\)
\(K=9A-5B\)
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)
1. \(A=\dfrac{2\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{9}-\dfrac{1}{11}\right)}{4\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{9}-\dfrac{1}{11}\right)}=\dfrac{2}{4}=\dfrac{1}{2}\)
2. \(B=\dfrac{1^2.2^2.3^2.4^2}{1.2^2.3^2.4^2.5}=\dfrac{1}{5}\)
3.\(C=\dfrac{2^2.3^2.\text{4^2.5^2}.5^2}{1.2^2.3^2.4^2.5.6^2}=\dfrac{125}{36}\)
4.D=\(D=\left(\dfrac{4}{5}-\dfrac{1}{6}\right).\dfrac{4}{9}.\dfrac{1}{16}=\dfrac{19}{30}.\dfrac{1}{36}=\dfrac{19}{1080}\)
Lời giải:
Đặt $P=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{2001}{3^{2001}}$
$3P=1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{2001}{3^{2000}}$
$3P-P=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2000}}-\frac{2001}{3^{2001}}$
$2P+\frac{2001}{3^{2001}}=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2000}}$
$3(2P+\frac{2001}{3^{2001}})=3+1+\frac{1}{3}+...+\frac{1}{3^{1999}}$
$3(2P+\frac{2001}{3^{2001}})- (2P+\frac{2001}{3^{2001}})=3-\frac{1}{3^{2000}}$
$2(2P+\frac{2001}{3^{2001}}) =3-\frac{1}{3^{2000}}$
$P=\frac{1}{4}(3-\frac{4005}{3^{2001}})< \frac{3}{4}< \frac{4}{5}$