\(A=\frac{2}{2}-\frac{2}{3}+\frac{2}{3}-\frac{2}{4}+\frac{2}{4}-\frac{2}{5}+\frac{2}{5}-\frac{2}{6}+.....+\frac{2}{99}-\frac{2}{100}\)

Ta tính các số âm và số dương giống nhau cộng lại có tổng bằng 0

\(\Rightarrow A=\frac{2}{2}-\frac{2}{100}\)

\(A=\frac{100}{100}-\frac{2}{100}=\frac{98}{100}=\frac{49}{50}\)

Đúng 100%

Đúng 100%

Đúng 100%

\(A=\frac{2}{2\cdot3}+\frac{2}{3\cdot4}+\frac{2}{4\cdot5}+....+\frac{2}{99\cdot100}\)

\(A:2=\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+....+\frac{1}{99\cdot100}\)

A : 2 = \(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{99}-\frac{1}{100}\)

\(A:2=\frac{1}{2}-\frac{1}{100}\)

\(A:2=\frac{49}{100}\)

       A  = \(\frac{49}{50}\)

\(A=1\cdot2^2+2\cdot3^2+...+2017\cdot2018^2\)

\(=2^2\left(2-1\right)+3^2\left(3-1\right)+\cdots+2018^2\left(2018-1\right)\)

\(=\left(2^3+3^3+\cdots+2018^3\right)-\left(2^2+3^2+\cdots+2018^2\right)\)

\(=\left(1^3+2^3+\cdots+2018^3\right)-\left(1^2+2^2+\cdots+2018^2\right)\)

\(=\left(1+2+\cdots+2018\right)^2-\frac{2018\left(2018+1\right)\left(2\cdot2018+1\right)}{6}\)

\(=\left(2018\cdot\frac{2019}{2}\right)^2-1009\cdot673\cdot4037\)

=\(\mathbf{4.147.324.330.132}\)

c) \(C=1.2+2.3+3.4+...+98.99\)

\(\Rightarrow3C=1.2\left(3-0\right)+2.3\left(4-1\right)+3.4\left(5-2\right)+...+98.99\left(100-97\right)\)

\(=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+....+98.99.100-97.98.99\)

\(=98.99.100\)

\(\Rightarrow C=\frac{98.99.100}{3}=323400\)

d) \(D=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{99.101}\)

\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\)

\(=1-\frac{1}{101}=\frac{100}{101}\)

\(A=\frac{1^2.2^2.3^2...1000^2}{1.2^2.3^2.4^2...1000^2.1001}=\frac{1}{1001}\)

\(=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)

\(=2\left(\frac{1}{2}-\frac{1}{100}\right)\)

\(=2.\frac{49}{100}\)

\(=\frac{49}{50}\)

= 2.(1/2.3 + 1/3.4 + ... + 1/99.100)

trong ngoac co cong thuc do, tim hieu di la lam dc

\(\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}...\frac{100^2}{100.101}\)

\(=\frac{1.1.2.2.3.3...100.100}{1.2.2.3.3.4.4...100.101}\)

\(=\frac{\left(1.2.3...100\right)\left(1.2.3...100\right)}{\left(1.2.3..100\right)\left(2.3.4...101\right)}=\frac{1}{101}\)

Ta có: Q=4 + 2² + 2³ + .... + 2²⁰

          2Q = 8 + 2³ + 2⁴ + ... + 2²¹

          2Q - Q = 2²¹ - 4 - 2² + 8

          Q = 2²¹