Question

[Đỗ Hàn Hoài Phương]

Tính tổng P=1/2×3+1/6×5+1/10×7+...+1/198×101.

P/s: Giúp mk câu này vs nhoa^-^

Answer 1

[QuocDat]

Đặt \(A=\frac{1}{2.6}+\frac{1}{6.10}+...+\frac{1}{194.198}\)

\(A=\frac{1}{4}\left(\frac{1}{2}-\frac{1}{6}+\frac{1}{6}-\frac{1}{10}+...+\frac{1}{194}-\frac{1}{198}\right)\)

\(A=\frac{1}{4}\left(\frac{1}{2}-\frac{1}{198}\right)\)

\(A=\frac{1}{4}.\frac{49}{99}\)

\(A=\frac{49}{396}\)

Đặt \(B=\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\)

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

\(B=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{101}\right)\)

\(B=\frac{1}{2}.\frac{98}{303}\)

\(B=\frac{49}{303}\)

Vậy P = A + B = \(\frac{49}{396}+\frac{49}{303}\) Bạn tự tính luôn nha máy tính mình hết pin rồi

Answer 2

[ST]

\(P=\frac{1}{2.3}+\frac{1}{6.5}+\frac{1}{10.7}+...+\frac{1}{198.101}\)

\(P=\frac{1}{2}\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\right)\)

\(4P=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+....+\frac{2}{99.101}\)

\(4P=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{99}-\frac{1}{101}\)

\(4P=1-\frac{1}{101}\)

\(4P=\frac{100}{101}\)

\(P=\frac{100}{101}:4\)

\(P=\frac{25}{101}\)