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Ta có \(I=\frac{11}{3}+\frac{17}{3^2}+...+\frac{605}{3^{100}}\left(1\right)\)
\(\Leftrightarrow3I=11+\frac{17}{3}+\frac{23}{3^2}+...+\frac{605}{3^{99}}\left(2\right)\)
Lấy \(\left(2\right)trừ\left(1\right)\)ta có
\(3I-I=11+\frac{6}{3}+\frac{6}{3^2}+...+\frac{6}{3^{99}}-\frac{605}{3^{100}}\)
\(\Leftrightarrow2I=11+6\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\right)-\frac{605}{3^{100}}\)
Xét \(A=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{100}}\left(3\right)\)
\(\Leftrightarrow3A=1+\frac{1}{3}+...+\frac{1}{3^{99}}\left(4\right)\)
Lấy\(\left(4\right)-\left(3\right)\)ta có
\(2A=1-\frac{1}{3^{100}}\)
\(\Leftrightarrow6A=3-\frac{1}{3^{99}}\)
Khi đó \(2I=11+3-\frac{1}{3^{99}}-\frac{605}{3^{100}}\)
\(\Leftrightarrow2I=14-\left(\frac{1}{3^{99}}+\frac{605}{3^{100}}\right)\)
Vì\(\frac{1}{3^{99}}+\frac{605}{3^{100}}>0\)
\(\Rightarrow2I< 14\)
\(\Leftrightarrow I< 7\left(đpcm\right)\)
a) \(A=\frac{4}{3}+\frac{7}{3^2}+\frac{10}{3^3}+...+\frac{301}{3^{100}}\)
\(\Rightarrow3A=4+\frac{7}{3}+\frac{10}{3^2}+...+\frac{301}{3^{100}}\)
\(\Rightarrow3A-A=\left(4+\frac{7}{3}+\frac{10}{3^2}+...+\frac{301}{3^{99}}\right)-\left(\frac{4}{3}+\frac{7}{3^2}+...+\frac{301}{3^{100}}\right)\)
\(\Rightarrow2A=4+1+\frac{1}{3}+...+\frac{1}{3^{98}}-\frac{301}{3^{100}}\)
Đặt \(F=1+\frac{1}{3}+...+\frac{1}{3^{98}}\)
\(\Rightarrow3F=3+1+...+\frac{1}{3^{97}}\)
\(\Rightarrow3F-F=\left(3+...+\frac{1}{3^{97}}\right)-\left(1+...+\frac{1}{3^{98}}\right)\)
\(\Rightarrow2F=3-\frac{1}{3^{98}}< 3\)
\(\Rightarrow F< \frac{3}{2}\)
\(\Rightarrow2A< 4+\frac{3}{2}\)
\(\Rightarrow2A< \frac{11}{2}\)
\(\Rightarrow A< \frac{11}{4}\left(đpcm\right)\)
2. \(B=\frac{11}{3}+\frac{17}{3^2}+\frac{23}{3^3}+...+\frac{605}{3^{100}}\)
\(\Rightarrow3B=11+\frac{17}{3}+\frac{23}{3^2}+...+\frac{605}{3^{99}}\)
\(\Rightarrow3B-B=\left(11+...+\frac{605}{3^{99}}\right)-\left(\frac{11}{3}+...+\frac{605}{3^{100}}\right)\)
\(\Rightarrow2B=11+2+\frac{2}{3}+...+\frac{2}{3^{98}}-\frac{605}{3^{100}}\)
Đặt \(D=2+\frac{2}{3}+...+\frac{2}{3^{98}}\)
\(\Rightarrow3D=6+2+...+\frac{2}{3^{97}}\)
\(\Rightarrow2D=6-\frac{2}{3^{98}}< 6\)( làm tắt )
\(\Rightarrow2D< 6\)
\(\Rightarrow D< 3\)
\(\Rightarrow2B< 11+3\)
\(\Rightarrow2B< 14\)
\(\Rightarrow B< 7\left(đpcm\right)\)
\(b)\) Đặt \(B=\frac{3}{10}+\frac{3}{11}+\frac{3}{12}+\frac{3}{13}+\frac{3}{14}\) ta có :
\(B>\frac{3}{15}+\frac{3}{15}+\frac{3}{15}+\frac{3}{15}+\frac{3}{15}=\frac{3+3+3+3+3}{15}=\frac{3.5}{15}=\frac{15}{15}=1\)
\(\Rightarrow\)\(B>1\) \(\left(1\right)\)
Lại có :
\(B< \frac{3}{10}+\frac{3}{10}+\frac{3}{10}+\frac{3}{10}+\frac{3}{10}=\frac{3+3+3+3+3}{10}=\frac{3.5}{10}=\frac{15}{10}< \frac{20}{10}=2\)
\(\Rightarrow\)\(B< 2\) \(\left(2\right)\)
Từ (1) và (2) suy ra :
\(1< B< 2\) ( đpcm )
Vậy \(1< B< 2\)
Chúc bạn học tốt ~
bài làm
C=1+3+32+.............+3100
C=3C−C2
3C=3+32+33+.............+399+3100+3101
C=1+3+32+..................+399+3100
3C-C=(3+32+33+.............+399+3100+3101)-(1+3+32+..................+399+3100)
Triệt tiêu các số hạng co giá trị tuyệt đối bằng nhau, ta được:
2C=-1+3100
⇒C=3100−12
D=2/D+D/3
2D=2101-2100+299-298+..............+23-22
D=2100-299+298-297+............+22-2
2D+D=2101-2100+299-298+..............+23-22+2100-299+298-297+............+22-2
Triệt tiêu các số hạng có giá trị tuyệt đối bằng nhau, ta được:
3D=2101-2
⇒D=2101−23
B=31×4 +54×9 +79×16 +.........+1981×100
Quan sát biểu thức, ta có nhận xét:
4-1=3;
9-4=5;
16-9=7;
.......;100-81=19
=> Hiệu hai số ở mẫu bằng giá trị ở tử
⇒B=1−14 +14 −19 +19 −116 +.......+181 −1100
⇒B=1−1/100
B=99/100 <100/100
Vậy B<1
a) \(\frac{-8}{3}+\frac{7}{5}+\frac{-71}{15}\)< \(x\) < \(\frac{-13}{7}+\frac{19}{14}+\frac{-7}{2}\)
Ta có: \(\frac{-8}{3}+\frac{7}{5}+\frac{-71}{15}\)
=\(\frac{-40}{15}+\frac{21}{15}+\frac{-71}{15}\)
=\(\frac{-90}{15}\)
=\(-6\)
Ta có: \(\frac{-13}{7}+\frac{19}{14}+\frac{-7}{2}\)
=\(\frac{-26}{14}+\frac{19}{14}+\frac{-49}{14}\)
=\(\frac{-56}{14}\)
=\(-4\)
=> \(-6\)< \(x\)<\(-4\)
=> \(x=-5\)
b)\(\frac{5}{17}+\frac{-4}{9}+\frac{-20}{31}+\frac{12}{17}+\frac{-11}{31}\)< \(\frac{x}{9}\)<\(\frac{-3}{7}+\frac{7}{15}+\frac{4}{-7}+\frac{8}{15}+\frac{2}{3}\)
Ta có: \(\frac{5}{17}+\frac{-4}{9}+\frac{-20}{31}+\frac{12}{17}+\frac{-11}{31}\)
=\(\left(\frac{5}{17}+\frac{12}{17}\right)+\left(\frac{-20}{31}+\frac{-11}{31}\right)+\frac{-4}{9}\)
=\(1+\left(-1\right)+\frac{-4}{9}\)
=\(0+\frac{-4}{9}\)
=\(\frac{-4}{9}\)
Ta có: \(\frac{-3}{7}+\frac{7}{15}+\frac{4}{-7}+\frac{8}{15}+\frac{2}{3}\)
=\(\frac{-3}{7}+\frac{7}{15}+\frac{-4}{7}+\frac{8}{15}+\frac{2}{3}\)
=\(\left(\frac{-3}{7}+\frac{-4}{7}\right)+\left(\frac{7}{15}+\frac{8}{15}\right)+\frac{2}{3}\)
=\(\left(-1\right)+1+\frac{2}{3}\)
=\(0+\frac{2}{3}\)
=\(\frac{2}{3}\)
=> \(\frac{-4}{9}\)< \(\frac{x}{9}\)<\(\frac{2}{3}\)
=
=> \(\frac{-4}{9}\)<\(\frac{x}{9}\)<\(\frac{6}{9}\)
=> \(-4\)< \(x\)<\(6\)
=>\(x\in\left\{-3;-2;-1;0;1;2;3;4;5\right\}\)
1/ Tính:
\(\frac{3}{2}-\frac{5}{6}+\frac{7}{12}-\frac{9}{20}+\frac{11}{30}-\frac{13}{42}+\frac{15}{56}-\frac{17}{72}+\frac{19}{90}\)
\(=\frac{3}{1.2}-\frac{5}{2.3}+\frac{7}{3.4}-\frac{9}{4.5}+\frac{11}{5.6}-\frac{13}{6.7}+\frac{15}{7.8}-\frac{17}{8.9}+\frac{19}{9.10}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(=1-\frac{1}{10}\)
\(=\frac{9}{10}\)
Ta có \(H=\frac{7}{3}+\frac{13}{3^2}+...+\frac{605}{3^{100}}\)
\(\Leftrightarrow3H=7+\frac{13}{3}+...+\frac{605}{3^{99}}\)
\(\Rightarrow2H=7+\frac{6}{3}+\frac{6}{3^2}+...+\frac{6}{3^{99}}-\frac{605}{3^{100}}\)
\(\Leftrightarrow2H=7+6\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\right)-\frac{605}{3^{100}}\)
Mà \(6\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\right)=3-\frac{1}{3^{99}}\)
\(\Rightarrow2H=7+3-\left(\frac{1}{3^{99}}+\frac{605}{3^{100}}\right)\)
\(\Leftrightarrow2H=10-\left(\frac{1}{3^{99}}+\frac{605}{3^{100}}\right)\)
Vì\(\frac{1}{3^{99}}+\frac{605}{3^{100}}>0\)
\(\Rightarrow2H< 10\)
\(\Leftrightarrow H< 5\left(1\right)\)
Ta có \(2H=10-\left(\frac{1}{3^{99}}+\frac{605}{3^{100}}\right)\)
Mà\(\frac{1}{3^{97}}+\frac{605}{3^{98}}< 22\)
hay\(\frac{1}{3^{99}}+\frac{605}{3^{98}}< \frac{22}{9}\)
\(\Rightarrow2H>10-\frac{22}{9}=\frac{68}{9}=2\cdot\left(3+\frac{7}{9}\right)\)
\(\Rightarrow H>3+\frac{7}{9}\left(2\right)\)
Từ \(\left(1\right)\left(2\right)\Rightarrowđpcm\)
Sai r