Question

B=10/1×3+10/3×5+...+10/101×103

C=1/2×4+1/4×6+...+1/2022×2024

D=3/1×3+3/3×5+...+3/99×101

Giải giúp mình vs ạ,mình cần gấp

Answer 1

Câu 1:

$B=\frac{10}{1.3}+\frac{10}{3.5}+\frac{10}{5.7}+...+\frac{10}{101.103}$

$B=5(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{101.103})$

$=5(\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+...+\frac{103-101}{101.103})$

$=5(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{101}-\frac{1}{103})$

$=5(1-\frac{1}{103})=5.\frac{102}{103}=\frac{510}{103}$

Answer 2

Câu 2:

\(C=\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+....+\frac{1}{2022.2024}\\ =\frac{1}{2}\left[\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+....+\frac{2}{2022.2024}\right]\)

\(=\frac{1}{2}\left[\frac{4-2}{2.4}+\frac{6-4}{4.6}+\frac{8-6}{6.8}+....+\frac{2024-2022}{2022.2024}\right]\)

\(=\frac{1}{2}(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2022}-\frac{1}{2024})\\ =\frac{1}{2}(\frac{1}{2}-\frac{1}{2024})=\frac{1011}{4048}\)

Answer 3

Câu 3:

\(D=\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{99.101}\\ =\frac{3}{2}(\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+...+\frac{101-99}{99.101})\)

\(=\frac{3}{2}(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101})\)

\(=\frac{3}{2}(1-\frac{1}{101})=\frac{150}{101}\)