CMR: Với mọi số tự nhiên n lớn hơn 2 thì \(\frac{\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2n}}{1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{2n-1}}< \frac{n}{n+1}\)
CMR với mọi số tự nhiên \(n\ge1\):
a ) \(\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{\left(2n\right)^2}< \frac{1}{2}\)
b ) \(\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{7^2}+...+\frac{1}{\left(2n+1\right)^2}< \frac{1}{4}.\)
a ) \(A=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{\left(2n\right)^2}=\frac{1}{4}\left(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}\right)\)
\(< \frac{1}{4}\left(1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{\left(n-1\right)n}\right)=\frac{1}{4}\left(1+\frac{1}{1}-\frac{1}{n}\right)< \frac{1}{2}\)
b )
\(B=\frac{1}{3^2}+\frac{1}{5^2}+...+\frac{1}{\left(2n+1\right)^2}< \frac{1}{3^2-1}+\frac{1}{5^2-1}+...+\frac{1}{\left(2n+1\right)^2-1}\)
\(=\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{2n\left(2n+2\right)}\)
\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-...+\frac{1}{2n}-\frac{1}{2n+2}\right)\)
\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{2n+2}\right)< \frac{1}{4}\).
CMR với mọi số tự nhiên n>1 luôn có:
\(\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+...+\frac{1}{2n}< \frac{3}{4}\)
CMR với mọi số tự nhiên n>2 thì :
a)\(\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{\left(2n\right)^2}\)<\(\frac{1}{2}\)
b)\(\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{7^2}+...+\frac{1}{\left(2n+1\right)^2}\)<\(\frac{1}{4}\)
c)\(\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)\left(1+\frac{1}{3.5}\right)...\left(1+\frac{1}{\left(2n+1\right)^2}\right)\)<2
cho \(A=\frac{7}{3}.\frac{37}{3^2}....\frac{6^{2n}+1}{3^{2n}}\)và \(B=\left(1+\frac{1}{3}\right)\left(1+\frac{1}{3^2}\right)...\left(1+\frac{1}{3^{2n}}\right)\)với n thuộc N
a) Chứng minh: 5A-2B là số tự nhiên
b) Chứng minh với mọi số tự nhiên n khác 0 thì 5A-2B chia hết cho 45
CMR: \(N=\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{\left(2n\right)^2}< \frac{1}{4}\)(n thuộc N , n lớn hơn bằng 2)
Ta có :
\(N=\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{\left(2n\right)^2}\)
\(N=\frac{1}{2^2}.\left(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}\right)\)
Ta thấy : \(\frac{1}{2^2}< \frac{1}{1.2}\)
\(\frac{1}{3^2}< \frac{1}{2.3}\)
.......
\(\frac{1}{n^2}< \frac{1}{\left(n-1\right).n}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{\left(n-1\right).n}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n-1}-\frac{1}{n}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}< 1-\frac{1}{n}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}< 1\)
\(\Rightarrow\frac{1}{2^2}.\left(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}\right)< 1.\frac{1}{2^2}\)
\(\Rightarrow N< \frac{1}{4}\)(ĐPCM)
Ủng hộ mk nha !!! ^_^
chứng minh rằng với mọi số tự nhiên n>=0 thì
\(S=\frac{1}{2^2}+\frac{1}{2^4}+\frac{1}{2^6}+....+\frac{1}{\left(2n\right)^2}<\frac{1}{2}\)
CMR : với mọi số tự nhiên n > 1, ta có :
\(S=\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n}>\frac{13}{24}\)
chứng minh bài toán theo cách quy nạp toán học.
Với n=2 suy ra:\(\frac{1}{3}+\frac{1}{4}>\frac{13}{14}\left(TM\right)\)
Giả sử bài toán trên đúng với mọi n=k,ta cần chứng minh nó đúng với n=k+1,tức là:
\(S_k=\frac{1}{k+2}+\frac{1}{k+3}+\frac{1}{k+4}+....+\frac{1}{2\left(k+1\right)}>\frac{13}{14}\)
Thật vậy:
\(\frac{1}{k+2}+\frac{1}{k+3}+...+\frac{1}{2\left(k+1\right)}\)
\(=\frac{1}{k+1}+\frac{1}{k+2}+....+\frac{1}{2k}+\frac{1}{2k+1}+\frac{1}{2k+2}-\frac{1}{k+1}\)
\(=S_k+\frac{1}{2k+1}+\frac{1}{2k+2}-\frac{1}{k+1}\)
\(>\frac{13}{14}+\frac{2k+2}{2\left(k+1\right)\left(2k+1\right)}+\frac{2k+1}{2\left(k+1\right)\left(2k+1\right)}-\frac{2\left(2k+1\right)}{2\left(k+1\right)\left(2k+1\right)}\)
\(=\frac{13}{14}+\frac{2\left(k+1\right)+2k+1-2\left(2k+1\right)}{2\left(k+1\right)\left(2k+1\right)}\)
để dễ hiểu,,mik xin viết thêm nha(không phải để kiếm điểm,có người nhờ nên mới thế này:))
\(\frac{13}{14}+\frac{2\left(k+1\right)+2k+1-2\left(2k+1\right)}{2\left(k+1\right)\left(2k+1\right)}\)
\(=\frac{13}{14}+\frac{1}{2\left(k+1\right)\left(2k+1\right)}>\frac{13}{14}\left(k>1\right)\)
\(\Rightarrow S_{k+1}>\frac{13}{14}\)
\(\Rightarrow S_k>\frac{13}{14}\)
Phép chứng minh hoàn tất_._
\(S=\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n}>\frac{13}{24}\left(1\right)\)
Với : \(n=2\), suy ra : \(\frac{1}{3}+\frac{1}{4}=\frac{7}{12}>\frac{13}{24}\left(TM\right)\)
Giả sử : (1) đúng với : \(n=k\left(k>1\right)\), tức là :
\(S_k=\frac{1}{k+1}+\frac{1}{k+2}+...+\frac{1}{2k}>\frac{13}{24}\)( giả thiết quy nạp )
Ta cần c/m : (1) đúng với : \(n=k+1\), tức là cần chứng minh :
\(S_{k+1}=\frac{1}{k+2}+\frac{1}{k+3}+...+\frac{1}{2k}+\frac{1}{2k+1}+\frac{1}{2\left(k+1\right)}>\frac{13}{24}\)
Thật vậy : \(S_{k+1}=\frac{1}{k+2}+\frac{1}{k+3}+...+\frac{1}{2k}+\frac{1}{2k+1}+\frac{1}{2\left(k+1\right)}\)
\(=\frac{1}{k+1}+\frac{1}{k+2}+\frac{1}{k+3}+...+\frac{1}{2k}+\frac{1}{2k+1}+\frac{1}{2k+2}-\frac{1}{k+1}\)
\(=S_k+\frac{1}{2k+1}+\frac{1}{2k+2}-\frac{1}{k+1}>\frac{13}{24}+\frac{1}{2k+1}+\frac{1}{2k+2}-\frac{1}{k+1}\)
\(=\frac{13}{24}+\frac{2k+2}{2\left(k+1\right)\left(2k+1\right)}+\frac{2k+1}{2\left(k+1\right)\left(2k+1\right)}-\frac{2\left(2k+1\right)}{2\left(k+1\right)\left(2k+1\right)}\)
\(=\frac{13}{24}+\frac{2\left(k+1\right)+2k+1-2\left(2k+1\right)}{2\left(k+1\right)\left(2k+1\right)}\)
\(=\frac{13}{24}+\frac{2k+2+2k+1-4k-2}{2\left(k+1\right)\left(2k+1\right)}\)
\(=\frac{13}{24}+\frac{1}{2\left(k+1\right)\left(2k+1\right)}>\frac{13}{24}\left(k>1\right)\)
Vậy : \(\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n}>\frac{13}{24}\)đúng với mọi \(n>1\)
CMR \(\forall n\in\)N* ta có
\(\left(1-\frac{1}{2}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\left(\frac{1}{5}-\frac{1}{6}\right)+...+\left(\frac{1}{2n-1}-\frac{1}{2n}\right)=\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n}\)
Chứng minh rằng với mọi số tự nhiên n>1
b)\(\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{7^2}+...+\frac{1}{\left(2n+1\right)^2}< \frac{1}{4}\)
Đặt \(A=\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{7^2}+...+\frac{1}{\left(2n+1\right)^2}\)
Ta có : \(\left(2n+1\right)^2=4n^2+4n+1>4n^2+4n\Leftrightarrow\left(2n+1\right)^2>2n\left(2n+2\right)\)\(\Leftrightarrow\frac{1}{\left(2n+1\right)^2}< \frac{1}{2n\left(2n+2\right)}\)
Mà \(\hept{\begin{cases}\frac{1}{3^2}< \frac{1}{2.4}\\\frac{1}{5^2}< \frac{1}{4.6}\\\frac{1}{7^2}< \frac{1}{6.8}\end{cases}}\)
\(...............\)
\(\frac{1}{\left(2n+1\right)^2}< \frac{1}{2n\left(2n+2\right)}\)
\(\Rightarrow\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{7^2}+...+\frac{1}{\left(2n+1\right)^2}< \frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{2n\left(2n+2\right)}=B\)
\(=\frac{4-2}{2.4}+\frac{6-4}{4.6}+\frac{8-6}{6.8}+...+\frac{2n+2-2n}{2n\left(2n+2\right)}\)
\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2n}-\frac{1}{2n+2}\)
\(=\frac{1}{2}-\frac{1}{2n+2}< \frac{1}{2}\Rightarrow B< \frac{1}{4}\)
\(\Rightarrow A< B< \frac{1}{4}\Rightarrow A< \frac{1}{4}\) hay đpcm