Question

[nhung]

3/4+8/9+15/16+...+9999//10000

Answer 1

[Phan Trần Bảo Ngọc]

\(\frac{3}{4}+\frac{8}{9}+\frac{15}{16}\)+...+\(\frac{9999}{10000}\)

= (1-\(\frac{1}{4}\)) +(1-\(\frac{1}{9}\))+(1-\(\frac{1}{16}\))+...+(1-\(\frac{1}{10000}\))

= 99 - (\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}\)+....+\(\frac{1}{100^2}\)) => 99 - A

 Dễ thấy A>0 =>S < 99 (1)

Lại có A= \(\frac{1}{2^2}\)+\(\frac{1}{3^2}\)+\(\frac{1}{4^2}\)+...+\(\frac{1}{100^2}\)

=> A<\(\frac{1}{1.2}\)+\(\frac{1}{2.3}\)+\(\frac{1}{3.4}\)+...+\(\frac{1}{99.100}\)

=>A<1-\(\frac{1}{2}\)+\(\frac{1}{2}\)-\(\frac{1}{3}\)+\(\frac{1}{3}\)-\(\frac{1}{4}\)+...\(\frac{1}{99}\)-\(\frac{1}{100}\)

=>A<1-\(\frac{1}{100}\)<1

...

Answer 2

[buivanloi]

Hzzz tự làm ko nổi