Question

[Nguyễn Sang]

a)\(\frac{1}{1\cdot4}\)+\(\frac{1}{4\cdot7}\)+\(\frac{1}{7\cdot10}\)+. . .+\(\frac{1}{100\cdot103}\)

b)\(\frac{1}{2000\cdot1999}\)-\(\frac{1}{1999\cdot1998}\)-\(\frac{1}{1998\cdot1997}\)-. . .-\(\frac{1}{3\cdot2}\)-\(\frac{1}{2\cdot1}\)

c) \(\frac{-1}{3}\)+\(\frac{-1}{15}\)+\(\frac{-1}{35}\)+\(\frac{-1}{63}\)+. . .+\(\frac{-1}{9999}\)

Mong mọi người giúp đỡ

Answer 1

[Nguyễn Thanh Hiền]

a) \(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{100.103}\)

\(=\frac{1}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{100.103}\right)\)

\(=\frac{1}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{100}-\frac{1}{103}\right)\)

\(=\frac{1}{3}.\left(1-\frac{1}{103}\right)\)

\(=\frac{1}{3}.\frac{102}{103}\)

\(=\frac{34}{103}\)

b) \(\frac{1}{2000.1999}-\frac{1}{1999.1998}-\frac{1}{1998.1997}-...-\frac{1}{3.2}-\frac{1}{2.1}\)

\(=\frac{1}{2000.1999}-\left(\frac{1}{1999.1998}+\frac{1}{1998.1997}+...+\frac{1}{3.2}+\frac{1}{2.1}\right)\)(*)

Đặt biểu thức trong ngoặc là A ta có :

\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{1997.1998}+\frac{1}{1998.1999}\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{1997}-\frac{1}{1998}+\frac{1}{1998}-\frac{1}{1999}\)

\(A=1-\frac{1}{1999}\)

\(A=\frac{1998}{1999}\)

Thay vào biểu thức (*) ta có :

\(\frac{1}{2000.1999}-\frac{1998}{1999}\)

\(=\frac{1}{3998000}-\frac{1998}{1999}\)

\(=\frac{-3995999}{3998000}\)

c) \(\frac{-1}{3}+\frac{-1}{15}+\frac{-1}{35}+\frac{-1}{63}+...+\frac{-1}{9999}\)

\(=\frac{-1}{1.3}+\frac{-1}{3.5}+\frac{-1}{5.7}+\frac{-1}{7.9}+...+\frac{-1}{99.101}\)

\(=\frac{-1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{99.101}\right)\)

\(=\frac{-1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+....+\frac{1}{99}-\frac{1}{101}\right)\)

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

\(=\frac{-1}{2}.\frac{100}{101}\)

\(=\frac{-50}{101}\)

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