Chứng minh rằng: \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{\left(2n\right)^2}< \frac{1}{2}\)
Giúp mình với
Chứng minh rằng với mọi số thự nhiên \(n\ge1\)thì:
\(\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{\left(2n\right)^2}< \frac{1}{2}\)
GIÚP MÌNH VỚI TÍ MÌNH ĐI HỌC RÔI
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:\(\frac{1}{6^2}< \frac{1}{5.6};\frac{1}{8^2}< \frac{1}{6.7};\frac{1}{10^2}< \frac{1}{7.8}\)
\(\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+..........+\frac{1}{\left(2n\right)^2}=\frac{1}{2^2}\left(\frac{1}{1^2}+\frac{1}{2^2}+..........+\frac{1}{n^2}\right)\)
\(< \frac{1}{2^2}\left(1+\frac{1}{1.2}+\frac{1}{2.3}+......+\frac{1}{\left(n-1\right)n}\right)=\frac{1}{4}\left(1+1-\frac{1}{2}+.....-\frac{1}{n}\right)=\frac{1}{4}.\left(\frac{2n-1}{n}\right)\)
\(=\frac{2n-1}{4n}< \frac{2n}{4n}=\frac{1}{2}\Rightarrow\frac{1}{2^2}+\frac{1}{4^2}+....+\frac{1}{\left(2n\right)^2}< \frac{1}{2}\)
Chứng Minh Rằng
\(\left(1+\frac{1}{2^1}\right)\left(1+\frac{1}{2^2}\right)\left(1+\frac{1}{2^3}\right)...\left(1+\frac{1}{2^{2020}}\right)< 3\)
Giúp mình với tối nay mình đi học rồi.
Chứng minh rằng: \(\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{\left(2n\right)^2}
đặt A=1/2^2+1/4^2+1/6^2+.....+1/(2n)^2
ta có :
A=1/2^2 +1/2^2(1/2^2+1/3^2+1/4^2+.....+1/n^2)
A<1/2^2+1/2^2(1/1.2+1/2.3+...+1/(n-1)n)
=1/2^2+1/2^2(1-1/2+1/2-1/3+....+1/(n-1)-1/n)
=1/2^2+1/2^2(1-1/n)
<1/2^2+1/2^2.1=1/2<3/4
vậy A<3/4
1) Cho a thỏa mãn: \(a^5-a^3+a=2\) Chứng minh rằng: \(a^6< 4\)
2) Chứng minh rằng: \(\frac{1^2}{1.3}+\frac{2^2}{3.5}+\frac{3^2}{5.7}+...+\frac{n^2}{\left(2n-1\right)\left(2n+1\right)}=\frac{n}{2}-\frac{n^2}{4n+2}\)
1/ Ta có:
\(a^5-a^3+a=2\)
Dễ thấy a = 0 không phải là nghiệm từ đó ta có:
\(a^6-a^4+a^2=2a\)
\(\Rightarrow2a=a^6+a^2-a^4\ge2a^4-a^4\ge a^4\)
\(\Rightarrow\hept{\begin{cases}2a\ge a^4\\a>0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}2\ge a^3\\a>0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}4\ge a^6\\a>0\end{cases}}\)
Dấu = không xảy ra
Vậy \(a^6< 4\)
Câu 2/
Câu hỏi của XPer Miner - Toán lớp 9 - Học toán với OnlineMath
Bạn tham khảo cách làm của bạn Alibabba nguyễn nha!!
bài tập:
cho B=\(\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+\left(\frac{1}{2}\right)^4+...+\left(\frac{1}{2}\right)^{2014}+\left(\frac{1}{2}\right)^{2015}\) . chứng minh rằng: B<1
giúp mình với nha........
Chứng minh rằng: \(\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{\left(2n\right)^2}
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
CHỨNG MINH RẰNG:
\(\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{\left(2n\right)^2}< \frac{1}{4}\)
Đặt \(A=\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{\left(2n\right)^2}\)
\(=\frac{1}{2^2}.\left(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}\right)\)
Có:
\(\frac{1}{2^2}< \frac{1}{1.2}\)
\(\frac{1}{3^2}< \frac{1}{2.3}\)
\(...\)
\(\frac{1}{n^2}< \frac{1}{\left(n-1\right)n}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{\left(n-1\right)n}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n-1}-\frac{1}{n}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}< 1-\frac{1}{n}< 1\)
\(\Rightarrow A< \frac{1}{2^2}.1=\frac{1}{4}\)
CHỨNG MINH RẰNG:
\(\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+\frac{1}{10^2}+...+\frac{1}{\left(2n^2\right)}< \frac{1}{4}\)
=>\(\frac{1}{2^2}\)x (\(\frac{1}{2^2}\)+\(\frac{1}{3^2}\)+\(\frac{1}{4^2}\)+...+\(\frac{1}{n^2}\))
Đặt A=\(\frac{1}{2^2}\)+\(\frac{1}{3^2}\)+\(\frac{1}{4^2}\)+...+\(\frac{1}{n^2}\)
Ta có:\(\frac{1}{2^2}\)<\(\frac{1}{1\cdot2}\)
\(\frac{1}{3^2}\)<\(\frac{1}{2\cdot3}\)
.........\(\frac{1}{n^2}\)<\(\frac{1}{\left(n-1\right)\cdot n}\)
=>\(\frac{1}{2^2}\)+\(\frac{1}{3^2}\)+...+\(\frac{1}{n^2}\)<\(\frac{1}{1\cdot2}\)+\(\frac{1}{2\cdot3}\)+...+\(\frac{1}{\left(n-1\right)\cdot n}\)
=>A<1-\(\frac{1}{2}\)+\(\frac{1}{2}\)-\(\frac{1}{3}\)+...+\(\frac{1}{n-1}\)--\(\frac{1}{n}\)
=>\(\frac{1}{2^2}\)*A<\(\frac{1}{2^2}\)(1--\(\frac{1}{n}\))
=>\(\frac{1}{2^2}\)*A<\(\frac{1}{4}\)(1--\(\frac{1}{n}\))
=>\(\frac{1}{2^2}\)+\(\frac{1}{4^2}\)+...+\(\frac{1}{\left(2n\right)^2}\)<\(\frac{1}{4}\)--\(\frac{1}{4n}\)<\(\frac{1}{4}\)
=>\(\frac{1}{2^2}\)+\(\frac{1}{4^2}\)+...+\(\frac{1}{\left(2n\right)^2}\)<\(\frac{1}{4}\)