Question

Tinh tổng:(1-1/3)+(1-1/15)+(1-1/35)+.........+(1-1/9999)

Answer 1

\(\left(1-\frac{1}{3}\right)+\left(1-\frac{1}{15}\right)+...+\left(1-\frac{1}{9999}\right)\)

= \(\left(1-\frac{1}{1.3}\right)+\left(1-\frac{1}{3.5}\right)+...+\left(1-\frac{1}{99.101}\right)\)(50 cặp)

= \(\left(1+1+1+...+1\right)-\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{99.101}\right)\)(50 số hạng 1)

= \(1.50-\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{99.101}\right)\) 

= \(50-\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\right)\)

= \(50-\frac{1}{2}.\left(1-\frac{1}{101}\right)\)

= \(50-\frac{1}{2}.\frac{100}{101}\)

= \(50-\frac{50}{101}\)

= \(\frac{5000}{101}\)