A = 1 + 1/2^2 + 1/3^2 + 1/4^2 +… + 100^2<2 20/10/2021 Bởi Maya A = 1 + 1/2^2 + 1/3^2 + 1/4^2 +… + 100^2<2
Đáp án: Giải thích các bước giải: $A=1+\dfrac{1}{2^{2}}+\dfrac{1}{3^{2}}+…+\dfrac{1}{100^{2}}$ $ $ Ta có: $\dfrac{1}{2^{2}}<\dfrac{1}{1.2}$ $;$ $\dfrac{1}{3^{2}}<\dfrac{1}{2.3}$ $;…;$ $\dfrac{1}{100^{2}}<\dfrac{1}{99.100}$ $ $ $⇒A=1+\dfrac{1}{2^{2}}+\dfrac{1}{3^{2}}+…+\dfrac{1}{100^{2}}<1+\dfrac{1}{1.2}+\dfrac{1}{2.3}+…+\dfrac{1}{99.100}$ $ $ $⇒A=1+\dfrac{1}{2^{2}}+\dfrac{1}{3^{2}}+…+\dfrac{1}{100^{2}}<1+1-\dfrac{1}{100}$ $ $ $⇒A=1+\dfrac{1}{2^{2}}+\dfrac{1}{3^{2}}+…+\dfrac{1}{100^{2}}<2-\dfrac{1}{100}<2$ $ $ $⇒A=1+\dfrac{1}{2^{2}}+\dfrac{1}{3^{2}}+…+\dfrac{1}{100^{2}}<2$ (đpcm) Bình luận
Ta có $A = 1 + \dfrac{1}{2^2} + \dfrac{1}{3^2} + \cdots + \dfrac{1}{100}^2$ $< 1 + \dfrac{1}{1.2} + \dfrac{1}{2.3} + \cdots + \dfrac{1}{99.100}$ $= 1 + 1 – \dfrac{1}{2} + \dfrac{1}{2} – \dfrac{1}{3} +\cdots + \dfrac{1}{99} – \dfrac{1}{100}$ $= 2 – \dfrac{1}{100} < 2$ Vậy $A < 2$. Bình luận
Đáp án:
Giải thích các bước giải:
$A=1+\dfrac{1}{2^{2}}+\dfrac{1}{3^{2}}+…+\dfrac{1}{100^{2}}$
$ $
Ta có: $\dfrac{1}{2^{2}}<\dfrac{1}{1.2}$ $;$ $\dfrac{1}{3^{2}}<\dfrac{1}{2.3}$ $;…;$ $\dfrac{1}{100^{2}}<\dfrac{1}{99.100}$
$ $
$⇒A=1+\dfrac{1}{2^{2}}+\dfrac{1}{3^{2}}+…+\dfrac{1}{100^{2}}<1+\dfrac{1}{1.2}+\dfrac{1}{2.3}+…+\dfrac{1}{99.100}$
$ $
$⇒A=1+\dfrac{1}{2^{2}}+\dfrac{1}{3^{2}}+…+\dfrac{1}{100^{2}}<1+1-\dfrac{1}{100}$
$ $
$⇒A=1+\dfrac{1}{2^{2}}+\dfrac{1}{3^{2}}+…+\dfrac{1}{100^{2}}<2-\dfrac{1}{100}<2$
$ $
$⇒A=1+\dfrac{1}{2^{2}}+\dfrac{1}{3^{2}}+…+\dfrac{1}{100^{2}}<2$ (đpcm)
Ta có
$A = 1 + \dfrac{1}{2^2} + \dfrac{1}{3^2} + \cdots + \dfrac{1}{100}^2$
$< 1 + \dfrac{1}{1.2} + \dfrac{1}{2.3} + \cdots + \dfrac{1}{99.100}$
$= 1 + 1 – \dfrac{1}{2} + \dfrac{1}{2} – \dfrac{1}{3} +\cdots + \dfrac{1}{99} – \dfrac{1}{100}$
$= 2 – \dfrac{1}{100} < 2$
Vậy $A < 2$.