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Bài 1 :
Ta có :
\(A=\frac{10^{17}+1}{10^{18}+1}=\frac{\left(10^{17}+1\right).10}{\left(10^{18}+1\right).10}=\frac{10^{18}+10}{10^{19}+10}\)
Mà : \(\frac{10^{18}+10}{10^{19}+10}>\frac{10^{18}+1}{10^{19}+1}\)
Mà \(A=\frac{10^{18}+10}{10^{19}+10}\)nên \(A>B\)
Vậy \(A>B\)
Bài 2 :
Ta có :
\(S=\frac{2013}{2014}+\frac{2014}{2015}+\frac{2015}{2016}+\frac{2016}{2013}\)
\(\Rightarrow S=\frac{2014-1}{2014}+\frac{2015-1}{2015}+\frac{2016-1}{2016}+\frac{2013+3}{2013}\)
\(\Rightarrow S=1-\frac{1}{2014}+1-\frac{1}{2015}+1-\frac{1}{2016}+1+\frac{3}{2013}\)
\(\Rightarrow S=4+\frac{3}{2013}-\left(\frac{1}{2014}+\frac{1}{2015}+\frac{1}{2016}\right)\)
Vì \(\frac{1}{2013}>\frac{1}{2014}>\frac{1}{2015}>\frac{1}{2016}\)nên \(\frac{3}{2013}-\left(\frac{1}{2014}+\frac{1}{2015}+\frac{1}{2016}\right)>0\)
Nên : \(M>4\)
Vậy \(M>4\)
Bài 3 :
Ta có :
\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+.......+\frac{1}{100^2}\)
Suy ra : \(A< \frac{1}{1.3}+\frac{1}{2.4}+\frac{1}{3.5}+....+\frac{1}{99.101}\)
\(\Rightarrow A< \frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{2.4}+......+\frac{2}{99.101}\right)\)
\(\Rightarrow A< \frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-......-\frac{1}{101}\right)\)
\(\Rightarrow A< \frac{1}{2}.\left[\left(1+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{99}\right)-\left(\frac{1}{3}+\frac{1}{4}+......+\frac{1}{101}\right)\right]\)
\(\Rightarrow A< \frac{1}{2}.\left(1+\frac{1}{2}-\frac{1}{100}-\frac{1}{101}\right)\)
\(\Rightarrow A< \frac{1}{2}.\left(1+\frac{1}{2}\right)\)
\(\Rightarrow A< \frac{3}{4}\)
Vậy \(A< \frac{3}{4}\)
Bài 4 :
\(a)A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+....+\frac{1}{2015.2017}\)
\(\Rightarrow A=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+.....+\frac{1}{2015.2017}\right)\)
\(\Rightarrow A=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.....+\frac{1}{2015}-\frac{1}{2017}\right)\)
\(\Rightarrow A=\frac{1}{2}.\left(1-\frac{1}{2017}\right)\)
\(\Rightarrow A=\frac{1}{2}.\frac{2016}{2017}\)
\(\Rightarrow A=\frac{1008}{2017}\)
Vậy \(A=\frac{1008}{2017}\)
\(b)\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+......+\frac{1}{x\left(x+2\right)}=\frac{1008}{2017}\)
\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+......+\frac{2}{x.\left(x+2\right)}=\frac{2016}{2017}\)
\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.....+\frac{1}{x}-\frac{1}{x+2}=\frac{2016}{2017}\)
\(1-\frac{1}{x+2}=\frac{2016}{2017}\)
\(\Rightarrow\frac{1}{x+2}=1-\frac{2016}{2017}\)
\(\Rightarrow\frac{1}{x+2}=\frac{1}{2017}\)
\(\Rightarrow x+2=2017\)
\(\Rightarrow x=2017-2=2015\)
Vậy \(x=2015\)
=\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+............+\frac{1}{18.19.20}\)
=\(\frac{2}{1.2.3.2}+\frac{2}{2.3.4.2}+............+\frac{2}{18.19.20.2}\)
=\(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}............+\frac{1}{18.19}-\frac{1}{19.20}\)
=\(\frac{1}{1.2}-\frac{1}{19.20}\)
=\(\frac{189}{380}\)
Bạn viết thêm số thứ 3 ở đầu dãy thì mới biết quy luật của dãy để tính chứ. Viết 2 số thế kia ai tính được :D
Bạn chỉ viết 2 số ở đầu dãy thì ko thể biết được quy luật của dãy. Bạn cần cho thêm 1 số nữa mới giải được chi tiết nhé!
\(N=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{9^2}\)
\(=\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+....+\frac{1}{9.9}\)
\(N\)bé hơn \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{8.9}=N_1\)
\(N_1=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{8.9}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-.........-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\)
\(=1-\frac{1}{9}\)
\(=\frac{8}{9}\) \((1)\)
\(N\)lớn hơn \(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+.....+\frac{1}{9.10}=N_2\)
\(\Rightarrow N_2=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+......+\frac{1}{9.10}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-.....-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)
\(=\frac{1}{2}-\frac{1}{10}\)
\(=\frac{5}{10}-\frac{1}{10}=\frac{2}{5}\) \((2)\)
Từ \((1)\)và \((2)\)suy ra ; \(\frac{2}{5}\)bé hơn N bé hơn \(\frac{8}{9}\)
Học tốt
Nhớ kết bạn với mình
1/2+1/3+1/4+...+1/63>1/31+1/31+...+1/31(62 số hạng 1/31)
hay 1/2+1/3+1/4+...+1/63>62 x 1/31
nên 1/2+1/3+1/4+...+1/63>2(dpcm)
Ta có : \(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\)\(=1+\frac{1}{2.2}+\frac{1}{3.3}+...+\frac{1}{50.50}\)
Vì \(\frac{1}{2.2}< \frac{1}{1.2};\frac{1}{3.3}< \frac{1}{2.3};..;\frac{1}{50.50}< \frac{1}{49.50}\)nên :
\(\Rightarrow\) \(1+\frac{1}{2.2}+\frac{1}{3.3}+...+\frac{1}{50.50}\)\(< 1+\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{49.50}\)
Ta có : \(1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\)
\(=1+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\right)\)
\(=1+\left(1-\frac{1}{50}\right)\)\(=1+\frac{49}{50}\)
Vì \(\frac{49}{50}< 1\)nên \(1+\frac{49}{50}< 2\)\(\Rightarrow\)\(1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}< 2\)
\(\Rightarrow\)\(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\)\(< 1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}< 2\)
\(\frac{1}{2^2}>\frac{1}{1.2}=1-\frac{1}{2}\)
\(\frac{1}{3^2}>\frac{1}{2.3}=\frac{1}{2}-\frac{1}{3}\)
\(....\)
\(\frac{1}{2015^2}>\frac{1}{2014.2015}=\frac{1}{2014}-\frac{1}{2015}\)
nên \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2015^2}>1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2004}-\frac{1}{2005}\)
\(=1-\frac{1}{2005}\)
vì \(1-\frac{1}{2005}< 1\)
=> ĐPCM