![](https://rs.olm.vn/images/avt/0.png?1311)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt biểu thức trên là A, ta có:
A = 22 + 42 + 62 + ... + 2100
4A = 24 + 26 + ... + 2102
4A - A = ( 22 - 24 ) + ( 26 + 26 ) + ... + 2102 - 2
3A = 2102 - 22
A = \(\frac{2^{102}-2^2}{3}\)
\(A=2^2+4^2+6^2+..+98^2+100^2\)
\(A=2^2\left(1^2+2^2+3^2+...+49^2+50^2\right)=4\cdot\frac{50\cdot\left(50+1\right)\left(2\cdot50+1\right)}{6}=171700\)
![](https://rs.olm.vn/images/avt/0.png?1311)
A=2^2+4^2+6^2+...+98^2+100^2=\(2^2\left(1+2^2+3^2+...+50^2\right)\))=\(2^2.m\)
sau đó bạn chỉ cần tính 1^2+2^2+3^2+...+49^2+50^2
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
B=(1-2-3+4)+(5-6-7+8)+...+(97-98-99+100)
B=0+0+..+0
B=0
C=2^100-(2^99+2^98+2^97+...+1)
đặt D=2^99+2^98+2^97+...+1
=>D=2^100-1
=>C=2^100-(2^100-1)=1
![](https://rs.olm.vn/images/avt/0.png?1311)
L=22+22.22+22.32+...+22.492+22.502=22.(1+22+32+...+492+502)
Đặt biểu thức trong dấu ngặc là A
A=1+2.(3-1)+3(4-1)+...+49(50-1)+50(51-1)=1+2.3-.2+3.4-3+...+49.50-49+50.51-50
A=1+(2.3+3.4+4.5+...+49.50+50.51)-(2+3+4+...+49+50)
Đặt B=2.3+3.4+4.5+...+49.50+50.51
3B=2.3.3+3.4.3+4.5.3+...+49.50.3+50.51.3=2.3.(4-1)+3.4(5-2)+4.5.(6-3)+...+49.50.(51-48)+50.51(52-49)
3B=-1.2.3+2.3.4-2.3.4+3.4.5-3.4.5+4.5.6-...-48.49.50+49.50.51-49.50.51+50.51.52=50.51.52-1.2.3 => B=(50.51.52-1.2.3)/3
Đặt C=2+3+4+...+49+50 đây là cấp số cộng áp dụng công thức tính tổng S của 1 cấp số cộng sẽ tính được C
=> L=22.A=22.(1+B-C)
Bạn tự làm nốt nhé
![](https://rs.olm.vn/images/avt/0.png?1311)
A=-1++(-1)+..+-(1) có 50 số -1
=>A=-1x50=-50
B=(1-2-3+4)+(5-6-7+8)+...+(97-98-99+100)
B=0+0+0+..+0
B=0
C=2^100-(2^99+2^98+...+1)
C=2^100-(2^100-1)
C=1
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(M=\dfrac{\dfrac{1}{99}+\dfrac{2}{98}+\dfrac{3}{97}+\dfrac{4}{96}+...+\dfrac{97}{3}+\dfrac{98}{2}+\dfrac{99}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{100}}\)
\(=\dfrac{\left(1+\dfrac{1}{99}\right)+\left(1+\dfrac{2}{98}\right)+\left(1+\dfrac{3}{97}\right)+\left(1+\dfrac{4}{96}\right)+...+\left(1+\dfrac{98}{2}\right)+1}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{100}}\)
\(=\dfrac{\dfrac{100}{99}+\dfrac{100}{98}+\dfrac{100}{97}+...+\dfrac{100}{1}+\dfrac{100}{2}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{100}}\)
=100
Ta có: \(N=\dfrac{92-\dfrac{1}{9}-\dfrac{2}{10}-\dfrac{3}{11}-...-\dfrac{90}{98}-\dfrac{91}{99}-\dfrac{92}{100}}{\dfrac{1}{45}+\dfrac{1}{50}+\dfrac{1}{55}+...+\dfrac{1}{495}+\dfrac{1}{500}}\)
\(=\dfrac{\left(1-\dfrac{1}{9}\right)+\left(1-\dfrac{2}{10}\right)+\left(1-\dfrac{3}{11}\right)+...+\left(1-\dfrac{90}{98}\right)+\left(1-\dfrac{91}{99}\right)+\left(1-\dfrac{92}{100}\right)}{\dfrac{1}{5}\left(\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{99}+\dfrac{1}{100}\right)}\)
\(=\dfrac{\dfrac{8}{9}+\dfrac{8}{10}+\dfrac{8}{11}+...+\dfrac{8}{99}+\dfrac{8}{100}}{\dfrac{1}{5}\left(\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{99}+\dfrac{1}{100}\right)}\)
\(=\dfrac{8}{\dfrac{1}{5}}=40\)
\(\Leftrightarrow\dfrac{M}{N}=\dfrac{100}{40}=\dfrac{5}{2}\)
ua sao khong thay