Question

\(CMR:\dfrac{1}{3}+\dfrac{2}{3^2}+\dfrac{3}{3^3}+...+\dfrac{3}{3^{2001}}< \dfrac{4}{5}\)

Answer 1

Lời giải:
Đặt $P=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{2001}{3^{2001}}$

$3P=1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{2001}{3^{2000}}$

$3P-P=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2000}}-\frac{2001}{3^{2001}}$

$2P+\frac{2001}{3^{2001}}=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2000}}$

$3(2P+\frac{2001}{3^{2001}})=3+1+\frac{1}{3}+...+\frac{1}{3^{1999}}$
$3(2P+\frac{2001}{3^{2001}})- (2P+\frac{2001}{3^{2001}})=3-\frac{1}{3^{2000}}$

$2(2P+\frac{2001}{3^{2001}}) =3-\frac{1}{3^{2000}}$

$P=\frac{1}{4}(3-\frac{4005}{3^{2001}})< \frac{3}{4}< \frac{4}{5}$