Câu hỏi của Kim Jun Hun

Question

Chứng minh rằng:

A=$\dfrac{1}{1^2}+\dfrac{1}{2^2}+...+\dfrac{1}{2017^2}< 2$

Quan Nguyen Minh · Accepted Answer

Ta có:$\dfrac{1}{1^2}$<$\dfrac{1}{1\times2}$;$\dfrac{1}{2^2}$<$\dfrac{1}{2\times3}$;...................;$\dfrac{1}{2017^2}$<$\dfrac{1}{2017\times2018}$

$\Rightarrow$$\dfrac{1}{1^2}$$+$$\dfrac{1}{2^2}$$+$..........$+$$\dfrac{1}{2017^2}$<$\dfrac{1}{1\times2}$$+$$\dfrac{1}{2\times3}$$+$.........$+$$\dfrac{1}{2017\times2018}$

$\Rightarrow$$\dfrac{1}{1^2}$$+$$\dfrac{1}{2^2}$$+$..........$+$$\dfrac{1}{2017^2}$<$\dfrac{1}{1}$$-$$\dfrac{1}{2}$$+$$\dfrac{1}{2}$$-$$\dfrac{1}{3}$$+$.........$+$$\dfrac{1}{2017}$$-$$\dfrac{1}{2018}$

$\Rightarrow$$\dfrac{1}{1^2}$$+$$\dfrac{1}{2^2}$$+$..........$+$$\dfrac{1}{2017^2}$<$\dfrac{1}{1}$$-$$\dfrac{1}{2018}$<2

$\Rightarrow$$\dfrac{1}{1^2}$$+$$\dfrac{1}{2^2}+......+\dfrac{1}{2017^2}$<2Câu trả lời: Ta có:$\dfrac{1}{1^2}$<$\dfrac{1}{1\times2}$;$\dfrac{1}{2^2}$<$\dfrac{1}{2\times3}$;...................;$\dfrac{1}{2017^2}$<$\dfrac{1}{2017\times2018}$

$\Rightarrow$$\dfrac{1}{1^2}$$+$$\dfrac{1}{2^2}$$+$..........$+$$\dfrac{1}{2017^2}$<$\dfrac{1}{1\times2}$$+$$\dfrac{1}{2\times3}$$+$.........$+$$\dfrac{1}{2017\times2018}$

$\Rightarrow$$\dfrac{1}{1^2}$$+$$\dfrac{1}{2^2}$$+$..........$+$$\dfrac{1}{2017^2}$<$\dfrac{1}{1}$$-$$\dfrac{1}{2}$$+$$\dfrac{1}{2}$$-$$\dfrac{1}{3}$$+$.........$+$$\dfrac{1}{2017}$$-$$\dfrac{1}{2018}$

$\Rightarrow$$\dfrac{1}{1^2}$$+$$\dfrac{1}{2^2}$$+$..........$+$$\dfrac{1}{2017^2}$<$\dfrac{1}{1}$$-$$\dfrac{1}{2018}$<2

$\Rightarrow$$\dfrac{1}{1^2}$$+$$\dfrac{1}{2^2}+......+\dfrac{1}{2017^2}$<2

Bài 2: Cộng, trừ số hữu tỉ

Khoá học trên OLM (olm.vn)

Khoá học trên OLM (olm.vn)