Question

Bài 1 : Tính

a , `5 + 5^2 + 5^3 + ... + 5^99`

b, `\frac{1}{1.5} + \frac{1}{5.9} + ... + \frac{1}{17.21}`

Answer 1

a, Giải:

Đặt A = \(5+5^2+5^3+...+5^{99}\)

5A = \(5^2+5^3+...+5^{100}\)

5A - A = ( \(5^2+5^3+...+5^{100}\)) - ( \(5+5^2+5^3+...+5^{99}\))

4A = \(5^{100}-5\)

A = \(\frac{5^{100}-5}{4}\)

b, Giải:

Đặt B = \(\frac{1}{1.5}+\frac{1}{5.9}+...+\frac{1}{17.21}\)

4B = 4. (\(\frac{1}{1.5}+\frac{1}{5.9}+...+\frac{1}{17.21}\))

4B = \(\frac{4}{1.5}+\frac{4}{5.9}+...+\frac{4}{17.21}\)

4B = \(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+...+\frac{1}{17}-\frac{1}{21}\)

4B = 1 - \(\frac{1}{21}\)

4B = 20/21

B = \(\frac{\frac{20}{21}}{4}\) = 5/21