Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a. Ta có: \(\frac{1}{2^2}\)< \(\frac{1}{1.3}\)
\(\frac{1}{4^2}\)< 1/(3.5)
1/(6^2) <1/(5.7)
...
1/(2n)^2 < 1/(2n-1)(2n+1)
=> 1/2^2 +1/4^2 + 1/6^2 +...+1/(2n)^2 < 1/(1.3) +...+1/(2n-1)(2n+1)
=> 2(1/2^2 +1/4^2 + 1/6^2 +...+1/(2n)^2) < (1/1 - 1/3 +1/3 - 1/5 + 1/5 - 1/7 +...+ 1/(2n-1) - 1/(2n+1)
=>2(1/2^2 +1/4^2 + 1/6^2 +...+1/(2n)^2) < 1 - 1/(2n+1) = 2n/(2n+1)
=> 1/2^2 +1/4^2 + 1/6^2 +...+1/(2n)^2 < 2n/(2n+1) . 1/2
Vì 2n/2n+1 < 1 => 2n/(2n+1) . 1/2 < 1/2
=> 1/2^2 +1/4^2 + 1/6^2 +...+1/(2n)^2 <1/2
Câu b tương tự
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
Đặt \(T=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{\left(2n\right)^2}\)
\(< \frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{\left(2n-1\right)n}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2n-1}-\frac{1}{n}\)
\(=\frac{1}{2}-\frac{1}{n}< \frac{1}{2}^{\left(đpcm\right)}\) (không chắc nha)
Đặt \(A=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{\left(2n\right)^2}\)
\(=\frac{1}{2^2}.\left(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}\right)\)
Ta có: \(\frac{1}{1}=\frac{1}{1},\frac{1}{2^2}< \frac{1}{1.2},\frac{1}{3^2}< \frac{1}{2.3},....,\frac{1}{n^2}< \frac{1}{\left(n-1\right).n}\)
=> \(A< \frac{1}{2^2}.\left[1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{\left(n-1\right)n}\right]\)
\(=\frac{1}{2^2}.\left(1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{n}-\frac{1}{n+1}\right)\)
\(=\frac{1}{2^2}.\left(2-\frac{1}{n+1}\right)=\frac{1}{2}-\frac{1}{4.\left(n+1\right)}\)
p/s: bài tớ ko bt đúng ko, nhưng tth bn làm vậy sẽ ko có quy luật, đoạn này
nếu cứ theo quy luật, tiếp tục sẽ ntn:\(\frac{1}{6^2}< \frac{1}{5.6};\frac{1}{8^2}< \frac{1}{6.7};\frac{1}{10^2}< \frac{1}{7.8}\)
\(\frac{150}{5.8}+\frac{150}{8.11}+\frac{150}{11.14}+.....+\frac{150}{47.50}\)
\(=50.\left(\frac{3}{5.8}+\frac{5}{8.11}+.....+\frac{3}{47.50}\right)\)
\(=50.\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+......+\frac{1}{47}-\frac{1}{50}\right)\)
\(=50.\left(\frac{1}{5}-\frac{1}{50}\right)\)
\(=50.\frac{9}{50}=9\)
Ta có:
\(1^4+\frac{1}{4}=\left(1^2-1+\frac{1}{2}\right)\left(1^2+1+\frac{1}{2}\right)=\frac{1}{2}.\left(2+\frac{1}{2}\right)\)
\(2^4+\frac{1}{4}=\left(2^2-2+\frac{1}{2}\right)\left(2^2+2+\frac{1}{2}\right)=\left(2+\frac{1}{2}\right).\left(6+\frac{1}{2}\right)\)
\(3^4+\frac{1}{4}=\left(3^2-3+\frac{1}{2}\right)\left(3^2+3+\frac{1}{2}\right)=\left(6+\frac{1}{2}\right).\left(12+\frac{1}{2}\right)\)
\(4^4+\frac{1}{4}=\left(4^2-4+\frac{1}{2}\right)\left(4^2+4+\frac{1}{2}\right)=\left(12+\frac{1}{2}\right).\left(20+\frac{1}{2}\right)\)
...
\(19^4+\frac{1}{4}=\left(19^2-19+\frac{1}{2}\right)\left(19^2+19+\frac{1}{2}\right)=\left(342+\frac{1}{2}\right).\left(380+\frac{1}{2}\right)\)
\(20^4+\frac{1}{4}=\left(20^2-20+\frac{1}{2}\right)\left(20^2+20+\frac{1}{2}\right)=\left(380+\frac{1}{2}\right).\left(420+\frac{1}{2}\right)\)
=> \(\frac{\left(1^4+\frac{1}{4}\right)\left(3^4+\frac{1}{4}\right)\left(5^4+\frac{1}{4}\right)...\left(19^4+\frac{1}{4}\right)}{\left(2^4+\frac{1}{4}\right)\left(4^4+\frac{1}{4}\right)\left(6^4+\frac{1}{4}\right)...\left(20^4+\frac{1}{4}\right)}\)
\(=\frac{\frac{1}{2}\left(2+\frac{1}{2}\right)\left(6+\frac{1}{2}\right)\left(12+\frac{1}{2}\right)...\left(342+\frac{1}{2}\right).\left(380+\frac{1}{2}\right)}{\left(2+\frac{1}{2}\right)\left(6+\frac{1}{2}\right)\left(12+\frac{1}{2}\right)\left(20+\frac{1}{2}\right)...\left(380+\frac{1}{2}\right).\left(420+\frac{1}{2}\right)}\)
\(=\frac{\frac{1}{2}}{420+\frac{1}{2}}=\frac{1}{841}\)
