Question

[Mangekyou sharingan]

Tính

S = 2/ -1 . 3 + 2 / -3 . 5 + .... + 2/-99 . 101

Answer 1

[응 우옌 민 후엔]

\(S=\frac{2}{-1.3}+\frac{2}{-3.5}+...+\frac{2}{-99.101}\)

\(S=\frac{3-\left(-1\right)}{-1.3}+\frac{5-\left(-3\right)}{-3.5}+...+\frac{101-\left(-99\right)}{-99.101}\)

\(S=\frac{3}{-1.3}-\frac{-1}{-1.3}+\frac{5}{-3.5}-\frac{-3}{-3.5}+...+\frac{101}{-99.101}-\frac{-99}{-99.101}\)
\(S=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\)

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

\(S=\frac{100}{101}\)

Answer 2

[Nguyễn Linh Chi]

\(S=-\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{99.101}\right)=-\left(\frac{3-1}{3.1}+\frac{5-3}{3.5}+...+\frac{101-99}{99.101}\right)\)

\(=-\left(\frac{3}{3.1}-\frac{1}{3.1}+\frac{5}{3.5}-\frac{3}{3.5}+...+\frac{101}{99.101}-\frac{99}{99.101}\right)\)

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

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