cho A=1/1^2+1/2^2+1/3^2+….+1/100^2 ss A voi 199/100 05/11/2021 Bởi Iris cho A=1/1^2+1/2^2+1/3^2+….+1/100^2 ss A voi 199/100
$A = \dfrac{1}{1^2}+\dfrac{1}{2^2}+….+\dfrac{1}{100^2}$ $ < \dfrac{1}{1} + \dfrac{1}{1.2}+\dfrac{1}{2.3}+….+\dfrac{1}{99.100}$ $ = 1+\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+….+\dfrac{1}{99}-\dfrac{1}{100}$ $ = 1+1-\dfrac{1}{100} = \dfrac{199}{100}$ Vậy $A < \dfrac{199}{100}$ Bình luận
$A = \dfrac{1}{1^2}+\dfrac{1}{2^2}+….+\dfrac{1}{100^2}$
$ < \dfrac{1}{1} + \dfrac{1}{1.2}+\dfrac{1}{2.3}+….+\dfrac{1}{99.100}$
$ = 1+\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+….+\dfrac{1}{99}-\dfrac{1}{100}$
$ = 1+1-\dfrac{1}{100} = \dfrac{199}{100}$
Vậy $A < \dfrac{199}{100}$
Ở dưới nha
Sorry bạn chữ mik hơi xấu (^,^)