Question

[~Nhii_Sem'pai^^]

Tính tổng :

A =5+10+15+ . . . . . + 2015+2020

B = 2/1x3 + 2/3x5 + 2/5x7 + . . . . . + 2/99x101

C = 1/2x4 + 1/4x6 + 1/6x8 + . . . . . + 1/98x100

               Giải nhanh giúp mk nha ! ^.^

Answer 1

[Dương Lam Hàng]

a) Số số hạng của dãy A là: (2020-5):2+1 = 404 (số)

    Tổng A là: (2020+5)x404:2=409050

b) \(B=\frac{2}{1\times3}+\frac{2}{3\times5}+....+\frac{2}{99\times101}\)

        \(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{99}-\frac{1}{101}\)

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

c) \(C=\frac{1}{2\times4}+\frac{1}{4\times6}+\frac{1}{6\times8}+...+\frac{1}{98\times100}\)

         \(=\frac{1}{2}\times\left(\frac{2}{2\times4}+\frac{2}{4\times6}+\frac{2}{6\times8}+....+\frac{2}{98\times100}\right)\)

           \(=\frac{1}{2}\times\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+....+\frac{1}{98}-\frac{1}{100}\right)\)

             \(=\frac{1}{2}\times\left(1-\frac{1}{100}\right)=\frac{1}{2}\times\frac{99}{100}=\frac{99}{200}\)

Vậy .....

Answer 2

[Ninh]

A = 5 + 10 + 15 + ... + 2015 + 2020

Số số hạng là : 404

A = 409050

\(B=\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{99\cdot101}\)

\(B=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)

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

\(C=\frac{1}{2\cdot4}+\frac{1}{4\cdot6}+\frac{1}{6\cdot8}+...+\frac{1}{98\cdot100}\)

\(C=\frac{1}{2}\cdot\left(\frac{1}{2}-\frac{1}{4}\right)+\frac{1}{2}\cdot\left(\frac{1}{4}-\frac{1}{6}\right)+\frac{1}{2}\cdot\left(\frac{1}{6}-\frac{1}{8}\right)+...+\frac{1}{2}\cdot\left(\frac{1}{98}-\frac{1}{100}\right)\)

\(C=\frac{1}{2}\cdot\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{98}-\frac{1}{100}\right)\)

\(C=\frac{1}{2}\cdot\left(\frac{1}{2}-\frac{1}{100}\right)=\frac{1}{2}\cdot\frac{49}{100}=\frac{49}{200}\)