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\(\frac{1}{2^1}+\frac{2}{3^2}+\frac{3}{4^3}+...+\frac{99}{100^{99}}<\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}=1-\frac{1}{100}<1\)
Vậy \(\frac{1}{2^1}+\frac{2}{3^2}+\frac{3}{4^3}+...+\frac{99}{100^{99}}<1\)
đặt \(A=\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}\)
\(\Rightarrow3A=1-\frac{2}{3}+\frac{3}{3^2}-\frac{4}{3^3}+...+\frac{99}{3^{98}}-\frac{100}{3^{99}}\)
\(\Rightarrow A+3A=\left(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}\right)+\left(1-\frac{2}{3}+\frac{3}{3^2}-\frac{4}{3^3}+...+\frac{99}{3^{98}}-\frac{100}{3^{99}}\right)\)
\(\Rightarrow4A=1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}-\frac{100}{3^{100}}\)<\(B=1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}\)
\(\Rightarrow3B=3-1+\frac{1}{3}-\frac{1}{3^2}+...+\frac{1}{3^{97}}-\frac{1}{3^{98}}\)
\(\Rightarrow B+3B=\left(1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}\right)+\left(3-1+\frac{1}{3}-\frac{1}{3^2}+...+\frac{1}{3^{97}}-\frac{1}{3^{98}}\right)\)
\(\Rightarrow4B=3-\frac{1}{3^{98}}<3\)
\(\Rightarrow B<\frac{3}{4}\Rightarrow4A<\frac{3}{4}\Rightarrow A<\frac{3}{16}\)
\(\RightarrowĐPCM\)
\(A=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{99}{100}\)
\(\Rightarrow A>\frac{1}{2}.\frac{2}{3}.\frac{4}{5}...\frac{98}{99}\)
\(\Rightarrow A^2>\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}...\frac{98}{99}.\frac{99}{100}\)
\(\Rightarrow A^2>\frac{1}{100}=\frac{1}{10^2}\)
Vậy \(A>\frac{1}{10}\)
\(A=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{9999}{10000}\)
\(\Rightarrow A>\frac{1}{2}.\frac{2}{3}.\frac{4}{5}...\frac{9998}{9999}\)
\(\Rightarrow A^2>\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}...\frac{9998}{9999}.\frac{9999}{10000}\)
\(\Rightarrow A^2>\frac{1}{10000}=\frac{1}{100^2}\)
\(VayA>\frac{1}{100}=B\)
#)Giải :
\(A=\frac{1}{3^1}-\frac{1}{3^2}+\frac{1}{3^3}-\frac{1}{3^4}+...+\frac{1}{3^{99}}-\frac{1}{3^{100}}\)
\(A=\frac{2}{9^1}+\frac{2}{9^2}+\frac{2}{9^3}+...+\frac{2}{9^{50}}\)
\(\Rightarrow2A=1+\frac{2}{9}+\frac{2}{9^1}+\frac{2}{9^2}+\frac{2}{9^3}+...+\frac{2}{9^{49}}\)
\(\Rightarrow2A-A=A=\left(1+\frac{2}{9}+\frac{2}{9^1}+\frac{2}{9^2}+\frac{2}{9^3}+...+\frac{2}{9^{49}}\right)-\left(\frac{2}{9^1}+\frac{2}{9^2}+\frac{2}{9^3}+...+\frac{2}{9^{50}}\right)\)
\(\Rightarrow A=1+\frac{2}{9}-\frac{2}{9^{50}}=\frac{11}{9}-\frac{2}{9^{50}}\)
Có lẽ đúng .........................
Đặt 1/3_2/3^2+3/3^3_.......+99/3^99_100/3^100<3/16=A
3A=1_2/3+3/3^2_4/3^3+....+99/3^98_100/3^99
Lấy 3A+A=4A=1_1/3+1/3^2_1/3^3+1/3^4_....._1/3^99_100/3^100
4A<1_1/3+1/3^2_1/3^3+1/3^4_...+1/3^98_1/3^99(1)
Đặt B=1_1/3+1/3^2_1/3^3_1/3^4_....+1/3^98_1/3^99
+
3B=2+1/3_1/3^2+1/3^3+......+1/3^97_1/3^98
4B=3_1/3^99<3 suy ra 4B<3 suy ra B<3/4(2)
Từ (1) và (2) suy ra 4A<3/4
Suy ra A<3/16
Đặt : \(A=\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{100^2}\)
Ta thấy :
\(\frac{1}{5^2}< \frac{1}{4.5}\)
\(\frac{1}{6^2}< \frac{1}{5.6}\)
\(\frac{1}{7^2}< \frac{1}{6.7}\)
\(.......................\)
\(\frac{1}{100^2}< \frac{1}{99.100}\)
\(\Rightarrow A=\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{100^2}< \frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{99.100}\)
\(\Rightarrow A=\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{100^2}< \frac{1}{4}-\frac{1}{100}=\frac{6}{25}\)
Vì \(\frac{1}{6}< \frac{6}{25}< \frac{1}{4}\)nên \(\frac{1}{6}< A< \frac{1}{4}\)hay \(\frac{1}{6}< \frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{100^2}< \frac{1}{4}\)
~ Hok tốt ~
Bài 1:
Đặt \(A=\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{100^2}\)
Ta có:
\(A< \frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{99.100}=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}=\frac{1}{4}-\frac{1}{100}< \frac{1}{4}\)
Ta có:
\(A>\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{100.101}=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{101}=\frac{1}{5}-\frac{1}{101}>\frac{1}{6}\)
\(\Rightarrow\frac{1}{6}< \frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{100^2}< \frac{1}{4}\left(\text{đ}pcm\right)\)
Bài 2:
\(a)\)Tách tổng A thành ba nhóm:
\(A=\left(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{30}\right)+\left(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{50}\right)+\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{70}\right)\)
\(A>\frac{1}{30}\cdot20+\frac{1}{50}\cdot20+\frac{1}{70}\cdot20=\frac{2}{3}+\frac{2}{5}+\frac{2}{7}=1\frac{37}{105}\)
\(A>1\frac{35}{105}=1\frac{1}{3}=\frac{4}{3}\left(\text{đ}pcm\right)\)
\(b)\)Tách tổng A thành sáu nhóm:
\(A=\left(\frac{1}{11}+...+\frac{1}{20}\right)+\left(\frac{1}{21}+...+\frac{1}{30}\right)+\left(\frac{1}{31}+...+\frac{1}{40}\right)+\left(\frac{1}{41}+...+\frac{1}{50}\right)\)\(+\left(\frac{1}{51}+...+\frac{1}{60}\right)+\left(\frac{1}{61}+...+\frac{1}{70}\right)\)
\(A< \frac{1}{11}\cdot10+\frac{1}{21}\cdot10+\frac{1}{31}\cdot10+\frac{1}{41}\cdot10+\frac{1}{51}\cdot10+\frac{1}{61}\cdot10\)
\(A< 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}=1+\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{6}\right)+\left(\frac{1}{4}+\frac{1}{5}\right)< 2+0,5=2,5\left(\text{đ}pcm\right)\)
#Sakura
MỚI LÀM LÚC TỐI,HÊN QUÁ:
\(A=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}}\)
\(3A=1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{100}{3^{99}}\)
\(2A=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
\(6A=3+1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}\)
\(4A=3-\left(\frac{101}{3^{99}}-\frac{100}{3^{100}}\right)\)
\(4A=3-\frac{203}{3^{100}}\)
\(A=\frac{3}{4}-\frac{203}{3^{100}\cdot4}< \frac{3}{4}\)