Câu hỏi của Hoàng Nguyễn

Question

tính nhanh 1/2^2 + 1/3^2 + 1/4^2 + 1/5^2 +.......+1/n^2 < 1

kaitovskudo · Accepted Answer

Ta có:$\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};....;\frac{1}{n^2}< \frac{1}{\left(n-1\right)n}$

=>$\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}< \frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{\left(n-1\right)n}$

$=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n-1}-\frac{1}{n}$

$=1-\frac{1}{n}< 1$

=>$\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}< 1$(đpcm)

Câu trả lời:

Ta có:$\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};....;\frac{1}{n^2}< \frac{1}{\left(n-1\right)n}$

=>$\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}< \frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{\left(n-1\right)n}$

$=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n-1}-\frac{1}{n}$

$=1-\frac{1}{n}< 1$

=>$\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}< 1$(đpcm)