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Đặt \(E=\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+...+\frac{1}{7^{99}}+\frac{1}{7^{100}}\)
\(\Rightarrow7E=1+\frac{1}{7}+\frac{1}{7^2}+...+\frac{1}{7^{98}}+\frac{1}{7^{99}}\)
\(\Rightarrow7E-E=\left(1+\frac{1}{7}+...+\frac{1}{7^{98}}+\frac{1}{7^{99}}\right)-\left(\frac{1}{7}+\frac{1}{7^2}+...+\frac{1}{7^{99}}+\frac{1}{7^{100}}\right)\)
\(\Rightarrow6E=1-\frac{1}{7^{100}}\)
\(\Rightarrow E=\frac{1-\frac{1}{7^{100}}}{6}\)
\(\Rightarrow A=\left(36-\frac{36}{7^{100}}\right):\frac{1-\frac{1}{7^{100}}}{6}\)
\(\Rightarrow A=36\left(1-\frac{1}{7^{100}}\right).\frac{6}{1-\frac{1}{7^{100}}}\)
\(\Rightarrow A=36.6=216\)
3n+2 - 2n+2 +3n - 2n = 3n . 32 - 2n. 22 +3n -2n
= 3n(32+1) - (2n.22 +2n)
=3n . 10 - 2n .5
=3n.10 - 2n-1 .2 .5
= 3n.10 - 2n-1 .10
= 10(3n - 2n-1)
vì 10 chia hết cho 10 nên 10(3n-2n-1) chia hết cho 10
=> 3n+2 - 2n+2 +3n -2n chia hết cho 10
Ai làm nhanh nhất mình sẽ **** xin cảm ơn các bạn mình đang cần gấp
Ta có\(\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{100.101}< \frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}< \frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}\)
\(\Rightarrow\frac{1}{5}-\frac{1}{101}< A< \frac{1}{4}-\frac{1}{100}\)(A là đề bài)
Mà \(\frac{1}{5}-\frac{1}{30}=\frac{1}{6}< \frac{1}{5}-\frac{1}{101}< A< \frac{1}{4}-\frac{1}{100}< \frac{1}{4}\)
\(\Rightarrow\frac{1}{6}< A< \frac{1}{4}\left(ĐPCM\right)\)
Ta có: \(\frac{1}{5\cdot6}< \frac{1}{5^2}=\frac{1}{5\cdot5}< \frac{1}{4\cdot5}\)
\(\frac{1}{6\cdot7}< \frac{1}{6^2}=\frac{1}{6\cdot6}< \frac{1}{5\cdot6}\)
\(\frac{1}{7\cdot8}< \frac{1}{7^2}=\frac{1}{7\cdot7}< \frac{1}{6\cdot7}\)
.............................
\(\frac{1}{100\cdot101}< \frac{1}{100^2}=\frac{1}{100\cdot100}< \frac{1}{99\cdot100}\)
Đặt \(A=\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+...+\frac{1}{100\cdot101}\)
\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{100}-\frac{1}{101}\)
\(=\frac{1}{5}-\frac{1}{101}=\frac{96}{505}>\frac{1}{6}\)
\(B=\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+...+\frac{1}{99\cdot100}\)
\(=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{4}-\frac{1}{100}< \frac{1}{4}\)
\(=>\frac{1}{6}< A< \frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{100^2}< B< \frac{1}{4}\)
\(=>\frac{1}{6}< \frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{100^2}< \frac{1}{4}\left(Đpcm\right)\)
#)Giải :
Bài 1 :
\(C=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\Leftrightarrow3C=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{100}}\)
\(\Leftrightarrow3C-C=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{100}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\right)\)
\(\Leftrightarrow2C=1-\frac{1}{3^{100}}\Leftrightarrow C=\frac{1-\frac{1}{3^{100}}}{2}< \frac{1}{2}\Rightarrow C< \frac{1}{2}\left(đpcm\right)\)
Bài 2 :
\(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}=\frac{3}{1.4}+\frac{5}{4.9}+\frac{7}{9.16}+...+\frac{19}{81.100}\)
\(=\left(1-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{9}\right)+\left(\frac{1}{9}-\frac{1}{16}\right)+...+\left(\frac{1}{81}-\frac{1}{100}\right)=1-\frac{1}{100}=\frac{99}{100}< 1\)
\(\Rightarrow\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}< 1\left(đpcm\right)\)
Đặt \(A=\frac{1}{7^2}-\frac{1}{7^4}+\frac{1}{7^6}+\frac{1}{7^8}+...+\frac{1}{7^{98}}-\frac{1}{7^{100}}\)
Nhân \(\frac{1}{7^2}\)vào A. Ta được:
\(A.\frac{1}{7^2}=\frac{1}{7^4}-\frac{1}{7^6}+\frac{1}{7^8}-...-\frac{1}{7^{98}}+\frac{1}{7^{100}}+\frac{1}{7^{102}}\)
\(A=\frac{1}{7^2}-\frac{1}{7^4}+\frac{1}{7^6}-\frac{1}{7^8}+...+\frac{1}{7^{98}}-\frac{1}{7^{100}}\)
Ta có: \(\frac{1}{7^2}.A+A=\frac{1}{49}-\frac{1}{7^{102}}\Rightarrow\frac{50}{49}.A=\frac{1}{49}-\frac{1}{7^{102}}\)
\(\Rightarrow A=\left(\frac{1}{49}-\frac{1}{7^{102}}\right)\frac{49}{50}< \frac{1}{5}^{\left(đpcm\right)}\)
a) A = \(\frac{1}{7^2}-\frac{1}{7^4}+\frac{1}{7^6}-\frac{1}{7^8}+...+\frac{1}{7^{98}}-\frac{1}{7^{100}}\)
Nhân \(\frac{1}{7^2}\)với A .Ta được :
A .\(\frac{1}{7^2}\)= \(\frac{1}{7^4}-\frac{1}{7^6}+\frac{1}{7^8}-...-\frac{1}{7^{98}}+\frac{1}{7^{100}}-\frac{1}{7^{102}}\)
Ta có : \(\frac{1}{7^2}.A+A=\frac{1}{49}-\frac{1}{7^{102}}\)
\(\Rightarrow\frac{50}{49}.A=\frac{1}{49}-\frac{1}{7^{102}}\)
\(\Rightarrow A.\left(\frac{1}{49}-\frac{1}{7^{102}}\right).\frac{49}{50}< \frac{1}{50}\left(đpcm\right)\)
b)Giả sử a1 >a2 > a3 ...> a2015 nên a1 > a2015
Theo đề ra ta có : \(\frac{1}{a_1}+\frac{1}{a_2}+...+\frac{1}{a_{2015}}< \frac{1}{2016}+\frac{1}{2015}+...+1=A\)
A< \(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{8}+\left(\frac{1}{8}+\frac{1}{8}+...+\frac{1}{8}\right)\)có 2007 số \(\frac{1}{8}\)
Mà \(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{8}+\left(\frac{1}{8}+\frac{1}{8}+...+\frac{1}{8}\right)< 1+1+...+\frac{2018}{8}\)
Giả sử trong 2015 số nguyên dương đã cho không có số nào bằng nhau .
Và a1 < a2 < a3 < ... < a2015
Ta có : \(\frac{1}{a_1}+\frac{1}{a_2}+\frac{1}{a_3}+...+\frac{1}{a_{2015}}\le1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}\)
\(\Rightarrow\frac{1}{a_1}+\frac{1}{a_2}+...+\frac{1}{a_{2011}}< 1+\frac{1}{2}+\frac{1}{2}+...+\frac{1}{2}=1+1007=1008\)
=> Giả sử là sai => ít nhất 2 trong 2015 số nguyên dương đã cho bằng nhau ( đpcm )
help me