Question

[Trần Bảo Hân]

Tính tổng:

\(B=\dfrac{5}{6}+\dfrac{5}{18}+\dfrac{5}{36}+\dfrac{5}{60}+...+\dfrac{5}{270}\)

Answer 1

[Nguyễn Ngọc Anh Minh]

\(B=\dfrac{5}{3}x\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{90}\right)=\)

\(=\dfrac{5}{3}x\left(\dfrac{1}{1x2}+\dfrac{1}{2x3}+\dfrac{1}{3x4}+\dfrac{1}{4x5}+...+\dfrac{1}{9x10}\right)\)

\(C=\dfrac{2-1}{1x2}+\dfrac{3-2}{2x3}+\dfrac{4-3}{3x4}+...+\dfrac{10-9}{9x10}=\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}=\)

\(=1-\dfrac{1}{10}=\dfrac{9}{10}\)

\(\Rightarrow B=\dfrac{5}{3}xC=\dfrac{5}{3}x\dfrac{9}{10}=\dfrac{3}{2}\)