CMR:G=1/4^2+1/6^2+1/8^2+...+1/(2n)^2<4
Mọi người giúp mình với. Mình cảm ơn
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
VN
Võ Ngọc Phương
VIP
23 tháng 7 2023
a) \(\dfrac{-5}{11}+\left(\dfrac{-6}{11}+1\right)\)
\(=\dfrac{-5}{11}+\left(\dfrac{-6}{11}+\dfrac{11}{11}\right)\)
\(=\dfrac{-5}{11}+\dfrac{5}{11}\)
\(=0\)
b) \(\dfrac{2}{3}+\left(\dfrac{5}{7}+\dfrac{-2}{3}\right)\)
\(=\dfrac{2}{3}+\dfrac{-2}{3}+\dfrac{5}{7}\)
\(=0+\dfrac{5}{7}\)
\(=\dfrac{5}{7}\)
c) \(\left(\dfrac{-1}{4}+\dfrac{5}{8}\right)+\dfrac{-3}{8}\)
\(=\dfrac{-1}{4}+\dfrac{-3}{8}+\dfrac{5}{8}\)
\(=\dfrac{-2}{8}+\dfrac{-3}{8}+\dfrac{5}{8}\)
\(=0\)
d) \(\dfrac{3}{4}.\dfrac{7}{25}+\dfrac{3}{4}.\dfrac{18}{25}\)
\(=\dfrac{3}{4}.\left(\dfrac{7}{25}+\dfrac{18}{25}\right)\)
\(=\dfrac{3}{4}.1\)
\(=\dfrac{3}{4}\)
Chúc bạn học tốt
ND
9 tháng 1 2018
(-1)^2n+(-1)^2n+1+(-1)^2n+2
= (-1)^2n+ (-1)^2n . (-1) +(-1)^2n . (-1)^2
=(-1)^2n . [-1+ (-1)+(-1)^2]
= 1 . 1
=1
4 tháng 6 2017
Ta có :3n chia hết cho n - 1
<=> 3n - 3 + 3 chia hết cho n - 1
<=> 3.(n - 1) + 3 chia hết cho n - 1
=> 3 chia hết cho n - 1
=> n - 1 thuộc Ư(3) = {-3;-1;1;3}
Ta có bảng :
| n - 1 | -3 | -1 | 1 | 3 |
| n | -2 | 0 | 2 | 4 |
4 tháng 6 2017
Ta có : 8 : n - 2
<=> n - 2 thuộc Ư(8) = {-8;-4;-2;-1;1;2;4;8}
Ta có bảng :
| n - 2 | -8 | -4 | -2 | -1 | 1 | 2 | 4 | 8 |
| n | -6 | -2 | 0 | 1 | 3 | 4 | 6 | 20 |
16 tháng 8 2023
a) \(\dfrac{13}{20}+\dfrac{3}{5}+x=\dfrac{5}{6}\)
\(\Rightarrow\dfrac{5}{4}+x=\dfrac{5}{6}\)
\(\Rightarrow x=\dfrac{5}{6}-\dfrac{5}{4}\)
\(\Rightarrow x=\dfrac{-5}{12}\)
b) \(x+\dfrac{1}{3}=\dfrac{2}{5}-\dfrac{-1}{3}\)
\(\Rightarrow x+\dfrac{1}{3}=\dfrac{11}{15}\)
\(\Rightarrow x=\dfrac{11}{15}-\dfrac{1}{3}\)
\(\Rightarrow x=\dfrac{2}{5}\)
c)\(\dfrac{-5}{8}-x=\dfrac{-3}{20}-\dfrac{-1}{6}\)
\(\dfrac{-5}{8}-x=\dfrac{1}{60}\)
\(\Rightarrow x=\dfrac{-5}{8}-\dfrac{1}{60}\)
\(\Rightarrow x=\dfrac{-77}{120}\)
d) \(\dfrac{3}{5}-x=\dfrac{1}{4}+\dfrac{7}{10}\)
\(\Rightarrow\dfrac{3}{5}-x=\dfrac{19}{20}\)
\(\Rightarrow x=\dfrac{3}{5}-\dfrac{19}{20}\)
\(\Rightarrow x=\dfrac{-7}{20}\)
e) \(\dfrac{-3}{7}-x=\dfrac{4}{5}+\dfrac{-2}{3}\)
\(\Rightarrow\dfrac{-3}{7}-x=\dfrac{2}{15}\)
\(\Rightarrow x=\dfrac{-3}{7}-\dfrac{2}{15}\)
\(\Rightarrow x=\dfrac{-59}{105}\)
16 tháng 8 2023
g) \(\dfrac{-5}{6}-x=\dfrac{7}{12}+\dfrac{-1}{3}\)
\(\Rightarrow\dfrac{-5}{6}-x=\dfrac{1}{4}\)
\(\Rightarrow x=\dfrac{-5}{6}-\dfrac{1}{4}\)
\(\Rightarrow x=\dfrac{-13}{12}\)
14 tháng 6 2023
`@` `\text {Ans}`
`\downarrow`
`a)`
\(\left(\dfrac{x}{2}-1\right)^3+2=-\dfrac{11}{8}\) phải k bạn nhỉ? `11/8` k có bậc lũy thừa nào `=5` á.
`=>`\(\left(\dfrac{x}{2}-1\right)^3=-\dfrac{11}{8}-2\)
`=>`\(\left(\dfrac{x}{2}-1\right)^3=-\dfrac{27}{8}\)
`=>`\(\left(\dfrac{x}{2}-1\right)^3=\left(-\dfrac{3}{2}\right)^3\)
`=>`\(\dfrac{x}{2}-1=-\dfrac{3}{2}\)
`=>`\(\dfrac{x}{2}=-\dfrac{3}{2}+1\)
`=>`\(\dfrac{x}{2}=-\dfrac{1}{2}\)
`=> x=1`
Vậy, `x=1`
`b)`
\(\left(\dfrac{x}{3}+\dfrac{1}{2}\right)\left(75\%-1\dfrac{1}{2}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{x}{3}+\dfrac{1}{2}=0\\0,75-1\dfrac{1}{2}x=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}\dfrac{x}{3}=-\dfrac{1}{2}\\-\dfrac{3}{2}x=\dfrac{75}{100}\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}2x=-3\\-3x\cdot100=2\cdot75\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=-\dfrac{3}{2}\\-3x\cdot100=150\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=-\dfrac{3}{2}\\-3x=1,5\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)
Vậy, `x={-3/2; -1/2}.`
23 tháng 7 2023
`@` `\text {Ans}`
`\downarrow`
`a)`
\(\left(\dfrac{7}{8}-\dfrac{3}{4}\right)\cdot1\dfrac{1}{3}-\dfrac{2}{3}\cdot0,5\)
`=`\(\dfrac{1}{8}\cdot\dfrac{4}{3}-\dfrac{1}{3}\)
`=`\(\dfrac{1}{6}-\dfrac{1}{3}=-\dfrac{1}{6}\)
`b)`
\(\left(2+\dfrac{5}{6}\right)\div1\dfrac{1}{5}+\left(-\dfrac{7}{12}\right)\)
`=`\(\dfrac{17}{6}\div1\dfrac{1}{5}-\dfrac{7}{12}\)
`=`\(\dfrac{85}{36}-\dfrac{7}{12}=\dfrac{16}{9}\)
`c)`
\(75\%-1\dfrac{1}{2}+0,5\div\dfrac{5}{12}\)
`=`\(-\dfrac{3}{4}+\dfrac{6}{5}=\dfrac{9}{20}\)
VN
Võ Ngọc Phương
VIP
23 tháng 7 2023
a) \(\left(\dfrac{7}{8}-\dfrac{3}{4}\right).1\dfrac{1}{3}-\dfrac{2}{3}.0,5\)
\(=\left(\dfrac{7}{8}-\dfrac{6}{8}\right).\dfrac{4}{3}-\dfrac{2}{3}.\dfrac{1}{2}\)
\(=\dfrac{1}{8}.\dfrac{4}{3}-\dfrac{2}{3}.\dfrac{1}{2}\)
\(=\dfrac{1}{6}-\dfrac{1}{3}\)
\(=\dfrac{-1}{6}\)
b) \(\left(2+\dfrac{5}{6}\right):1\dfrac{1}{5}+\dfrac{-7}{12}\)
\(=\left(\dfrac{12}{6}+\dfrac{5}{6}\right):\dfrac{6}{5}+\dfrac{-7}{12}\)
\(=\dfrac{17}{6}.\dfrac{5}{6}+\dfrac{-7}{12}\)
\(=\dfrac{85}{36}+\dfrac{-7}{12}\)
\(=\dfrac{16}{9}\)
c) \(75\%-1\dfrac{1}{2}+0,5:\dfrac{5}{12}\)
\(=\dfrac{3}{4}-\dfrac{3}{2}+\dfrac{1}{2}.\dfrac{12}{5}\)
\(=\dfrac{3}{4}-\dfrac{6}{4}+\dfrac{6}{5}\)
\(=\dfrac{-3}{4}+\dfrac{6}{5}\)
\(=\dfrac{9}{20}\)
\(G\)\(=\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{\left(2n\right)^2}\)
\(G=\frac{1}{4}\left(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}\right)\)
Đặt S = \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}\)
Ta thấy : \(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};......;\frac{1}{n^2}< \frac{1}{\left(n-1\right).n}\)
=> S < \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{\left(n-1\right).n}\)
=> S <\(1-\frac{1}{n}\)
Thay S vào G ta có :
G < \(\frac{1}{4}\left(1-\frac{1}{n}\right)\)
G< \(\frac{1}{4}-\frac{1}{4n}< \frac{1}{4}\)( đpcm )
Học tốt
#Dương