Ta có: (a - b).(a+ b) = a.(a+ b) - b.(a+ b) = a² + ab - ab  - b² = a² - b²

Áp dụng: M = (100 - 99)(100 + 99) + (98 - 97).(98 + 97) + ...+ (2 - 1)(2+1)

= 100 + 99 + 98 + 97 + ...+ 2 + 1 = (1+100).100 : 2 = 5050

Ta thấy: a^2 - (a-1)^2 = a^2 - (a-1) x (a-1) = a^2 - a(a-1) + (a-1) = a^2 - a^2 + a+a-1=2a-1

Áp dụng công thức này được:

 \(M=\left(100^2-99^2\right)+\left(98^2-97^2\right)+...+\left(2^2-1^2\right)\)

\(M=2.100-1+2.98-1+...+2.2-1\)

\(M=2\left(100+98+...+2\right)-50\)

\(M=2.\frac{\left[\left(100-2\right):2+1\right]\left(100+2\right)}{2}-50\)

\(M=50.102-50=50\left(102-1\right)=50.101=5050\)

5050 la dung day.

a)

C = 1 − 2 + 3 − 4 + ... + 97 − 98 + 99 − 100 = 1 − 2 + 3 − 4 + ... + 97 − 98 + 99 − 100 = − 1 + − 1 + ... + − 1 + − 1 = − 1.50 = − 50.

b)

B = 1 − 2 − 3 + 4 + 5 − 6 − 7 + ... + 97 − 98 − 99 + 100 = 1 − 2 + − 3 + 4 + 5 − 6 + ... + 97 − 98 + − 99 + 100 = − 1 + 1 + − 1 + ... + − 1 + 1 = − 1 + 1 + − 1 + 1 + ... + − 1 + 1 − 1 = 0 + 0 + ... + 0 − 1 = − 1.

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: \(N=\dfrac{92-\dfrac{1}{9}-\dfrac{2}{10}-\dfrac{3}{11}-...-\dfrac{90}{98}-\dfrac{91}{99}-\dfrac{92}{100}}{\dfrac{1}{45}+\dfrac{1}{50}+\dfrac{1}{55}+...+\dfrac{1}{495}+\dfrac{1}{500}}\)

\(=\dfrac{\left(1-\dfrac{1}{9}\right)+\left(1-\dfrac{2}{10}\right)+\left(1-\dfrac{3}{11}\right)+...+\left(1-\dfrac{90}{98}\right)+\left(1-\dfrac{91}{99}\right)+\left(1-\dfrac{92}{100}\right)}{\dfrac{1}{5}\left(\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{99}+\dfrac{1}{100}\right)}\)

\(=\dfrac{\dfrac{8}{9}+\dfrac{8}{10}+\dfrac{8}{11}+...+\dfrac{8}{99}+\dfrac{8}{100}}{\dfrac{1}{5}\left(\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{99}+\dfrac{1}{100}\right)}\)

\(=\dfrac{8}{\dfrac{1}{5}}=40\)

\(\Leftrightarrow\dfrac{M}{N}=\dfrac{100}{40}=\dfrac{5}{2}\)

gọi là A đi

=> 2A=\(2^2+2^3+...+2^{101}\)

=> \(2A-A=A=\left(2^2+2^3+...+2^{101}\right)-\left(2+2^2+...+2^{100}\right)=2^{101}-2=2\left(2^{100}-1\right)\)

ng thi bich hau co the giai chi tiet hon dc ko

\(=2^{100}-\left(2^{99}+2^{98}+2^{97}+...+2+1\right)\)

Đặt \(B=1+2+2^2+...+2^{98}+2^{99}\)

\(\Rightarrow2B=2+2^2+2^3+...+2^{100}\)

\(\Rightarrow B=\left(2+2^2+2^3+..+2^{100}\right)-\left(1+2+2^2+...+2^{99}\right)\)

\(\Rightarrow B=2^{100}-1\)

\(\Rightarrow2^{100}-2^{99}-2^{98}-....-2-1=2^{100}-\left(2^{100}-1\right)\)

\(=1\)

C = 1 − 2 + 3 − 4 + ... + 97 − 98 + 99 − 100 = 1 − 2 + 3 − 4 + ... + 97 − 98 + 99 − 100 = − 1 + − 1 + ... + − 1 + − 1 = − 1.50 = − 50.

Số số hạng là : 

\(\left(101-2\right):1+1=100\)

Tổng trên có giá trị là : 

\(\dfrac{\left(101+2\right).100}{2}=5150\)

A= 2 + 3+4+...+96+97+98+99+100+101

Khoảng cách của dãy số trên là: 3-2 =1

Số số hạng của dãy số trên là: (101 - 2): 1 + 1 = 100 (số hạng)

Tổng A là: A = (101+2)\(\times\) 100 : 2  =5150

Đáp số: 5150