Question

Chứng minh:

\(\dfrac{1}{3^2}+\dfrac{1}{4^2}+\dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}và< \dfrac{1}{2}\)

Answer 1

Ta có :

\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)

\(\dfrac{1}{4^2}< \dfrac{1}{3.4}\)

...................

\(\dfrac{1}{100^2}< \dfrac{1}{99.100}\)

\(\Leftrightarrow\dfrac{1}{3^2}+\dfrac{1}{4^2}+..........+\dfrac{1}{100^2}< \dfrac{1}{2.3}+\dfrac{1}{3.4}+.....+\dfrac{1}{99.100}\)

\(\Leftrightarrow\dfrac{1}{3^2}+\dfrac{1}{4^2}+....+\dfrac{1}{100^2}< \dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+......+\dfrac{1}{99}-\dfrac{1}{100}\)

\(\Leftrightarrow\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}< \dfrac{1}{2}-\dfrac{1}{100}\)

\(\Leftrightarrow\dfrac{1}{3^2}+\dfrac{1}{4^2}+.......+\dfrac{1}{100^2}< \dfrac{49}{100}< \dfrac{1}{2}\)

Answer 2

Tham khảo : https://olm.vn/hoi-dap/question/564353.html

Answer 3

đặt dãy tính trên là A,

ta có 1/ 3² = 1/3.3 < 1/ 2.3

1/4² = 1/4.4 < 1/ 3.4

1/5²= 1/5.5 < 1/ 4.5

1/6² = 1/6.6 < 1/ 5.6

..........

1/100² = 1/100. 100 < 1/ 99.100

Đặt D = 1/ 2.3 + 1/3.4 + 1/4.5 + 1/ 5.6+ ... + 1/ 99.100

suy ra A < D

mà D = 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + 1/5 - 1/6 + ....+ 1/99 - 1/ 100

= 1/2 - 1/100

Mà 1/2 - 1/100 < 1/2

hay D < 1/2 , suy ra A < D < 1/2

suy ra A < 1/2

vậy dãy tính trên < 1/2