Lời giải:
$A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{20}}$
$3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{19}}$
$\Rightarrow 3A-A=1-\frac{1}{3^{20}}$
$\Rightarrow A=\frac{1}{2}(1-\frac{1}{3^{20}})$
1 CMR:
B=\(\frac{4}{3}+\frac{7}{3^2}+\frac{10}{3^3}+.....+\frac{3n+1}{3^n}< \frac{11}{4}\)(n thuộc N*;n>3)
A=\(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}}< \frac{3}{4}\)
C=\(\frac{2}{3}+\frac{8}{9}+\frac{26}{27}+...+\frac{3^{20}-1}{3^{20}}>19\frac{1}{2}\)
Tính
a. S= 3 + \(\frac{3}{2}+\frac{3}{2^2}+......+\frac{3}{2^9}\)
b. A= \(\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}+\frac{1}{132}+\frac{1}{156}\)
c.M=\(\frac{20}{112}+\frac{20}{280}+\frac{20}{520}+\frac{20}{832}\)
A=\(\frac{2}{1+2}+\frac{2+3}{1+2+3}+\frac{2+3+4}{1+2+3+4}+...+\frac{2+3+...+20}{1+2+3+...+20}\)
Tính A=\(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{18}+\frac{1}{19}+\frac{1}{20}}{\frac{19}{1}+\frac{18}{2}+\frac{17}{3}+...+\frac{3}{17}+\frac{2}{18}+\frac{1}{19}}\)
bài 1: tính nhanh
a, \(\frac{\frac{1}{2}-\frac{1}{3}-\frac{1}{4}}{1-\frac{2}{3}-\frac{1}{2}}\)-\(\frac{\frac{3}{5}-\frac{3}{7}-\frac{3}{11}}{\frac{6}{5}-\frac{6}{7}-\frac{6}{11}}\)
b,\(1\frac{1}{2}+\frac{1}{2}+\frac{1}{4}+\frac{3}{20}+...+\frac{3}{2011.2012}\)
\(\frac{2}{1+2}+\frac{2+3}{1+2+3}+\frac{2+3+4}{1+2+3+4}+......+\frac{2+3+4+...+20}{1+2+3+4+...+20}\)
1 . Tinh : a , \(\left(1-\frac{1}{6}\right)\left(1-\frac{1}{10}\right)\left(1-\frac{1}{14}\right).....\left(1-\frac{1}{5050}\right)\)b,\(\frac{^{2^{19}}.\left(3^3\right)^3+3.5.\left(2^2\right)^9.\left(3^2\right)^4}{\left(2.3\right)^9.2^{10}+\left(3.2^2\right)^{10}}\)c,\(\frac{18.\frac{19}{2}.\frac{20}{3}.\frac{21}{4}.....\frac{36}{19}}{20.\frac{21}{2}.\frac{22}{3}.....\frac{36}{17}}\)giup mjk nha mjk tjk cho
\(Tính:\)
\(A=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{56}\)
\(B=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{11.13}\)
\(C=\frac{3}{4}+\frac{3}{28}+\frac{3}{70}+\frac{3}{130}+\frac{3}{208}+\frac{3}{304}\)
\(D=\frac{1}{2}+\frac{1}{14}+\frac{1}{35}+\frac{1}{65}+\frac{1}{104}+\frac{1}{152}\)
Tính
A=\(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{20}}{\frac{19}{1}+\frac{18}{2}+\frac{17}{3}+...+\frac{1}{19}}\)
Rút gọn:
A =\(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{20}\right)\)
B = \(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2012}}\)
So sánh :
\(A=\frac{20^{10}+1}{20^{10}-1}vàB=\frac{20^{10}-1}{20^{10}-3}\)
