Question

Bài 1: Cho:

A=\(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}\)

Chứng mình rằng A < 2

TICK CHO AI TRẢ LỜI ĐÚNG NHẤT

Answer 1

cứ mỗi p/số kia bé hơn:1+1/1.2+1/2.3+1/3.4+....+1/49.50

phân phối ra nhé còn:2-1/50

mà 1/50>0

=>A<2

Answer 2

A=\(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+.....+\frac{1}{50^2}\)

A=\(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}<\frac{1}{1.1}+\frac{1}{1.2}+....+\frac{1}{49.50}\)

A=\(\frac{1}{1}-\frac{1}{50}=\frac{50}{50}-\frac{1}{50}=\frac{49}{50}<2=\frac{2}{1}\)

A=\(\frac{49}{50}<\frac{2}{1}=\frac{49}{50}<\frac{100}{50}\)

Vậy A<2 hay\(\frac{49}{50}<2\)

Answer 3

A=\(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{50^2}\)

A=\(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{50^2}<\frac{1}{0.1}+\frac{1}{1.2}+...+\frac{1}{49.50}\)

A=\(\frac{1}{1}-\frac{1}{50}=\frac{50}{50}-\frac{1}{50}=\frac{49}{50}\)

A=\(\frac{49}{50}<2=\frac{49}{50}<\frac{100}{50}\)

Vậy A<2 hay \(\frac{49}{50}<\frac{2}{1}\)

Answer 4

dang ay cau tra loi roi ma chang dc cau nao hay sao ma chua k cau cua ban Sawada Tsunayoshi la dung roi hoi minh thay nguoi ta giai vay do