Chứng minh rằng: 1+1/2^2+1/3^2+1/4^2+…+1/100^2<2 27/10/2021 Bởi Margaret Chứng minh rằng: 1+1/2^2+1/3^2+1/4^2+…+1/100^2<2
Đáp án: Giải thích các bước giải: $1+\dfrac{1}{2^{2}}+\dfrac{1}{3^{2}}+…+\dfrac{1}{100^{2}}$ $ $ Ta có: $1=1$ ; $\dfrac{1}{2^{2}}<\dfrac{1}{1.2}$ ; $\dfrac{1}{3^{2}}<\dfrac{1}{2.3}$ ;…; $\dfrac{1}{100^{2}}<\dfrac{1}{99.100}$ $ $ $⇒1+\dfrac{1}{2^{2}}+\dfrac{1}{3^{2}}+…+\dfrac{1}{100^{2}}<1+\dfrac{1}{1.2}+\dfrac{1}{2.3}+…+\dfrac{1}{99.100}$ $ $ $⇒1+\dfrac{1}{2^{2}}+\dfrac{1}{3^{2}}+…+\dfrac{1}{100^{2}}<1+1-\dfrac{1}{100}$ $ $ $⇒1+\dfrac{1}{2^{2}}+\dfrac{1}{3^{2}}+…+\dfrac{1}{100^{2}}<2-\dfrac{1}{100}<2$ $ $ $⇒1+\dfrac{1}{2^{2}}+\dfrac{1}{3^{2}}+…+\dfrac{1}{100^{2}}<2$ Bình luận
Đáp án:
Giải thích các bước giải:
$1+\dfrac{1}{2^{2}}+\dfrac{1}{3^{2}}+…+\dfrac{1}{100^{2}}$
$ $
Ta có: $1=1$ ; $\dfrac{1}{2^{2}}<\dfrac{1}{1.2}$ ; $\dfrac{1}{3^{2}}<\dfrac{1}{2.3}$ ;…; $\dfrac{1}{100^{2}}<\dfrac{1}{99.100}$
$ $
$⇒1+\dfrac{1}{2^{2}}+\dfrac{1}{3^{2}}+…+\dfrac{1}{100^{2}}<1+\dfrac{1}{1.2}+\dfrac{1}{2.3}+…+\dfrac{1}{99.100}$
$ $
$⇒1+\dfrac{1}{2^{2}}+\dfrac{1}{3^{2}}+…+\dfrac{1}{100^{2}}<1+1-\dfrac{1}{100}$
$ $
$⇒1+\dfrac{1}{2^{2}}+\dfrac{1}{3^{2}}+…+\dfrac{1}{100^{2}}<2-\dfrac{1}{100}<2$
$ $
$⇒1+\dfrac{1}{2^{2}}+\dfrac{1}{3^{2}}+…+\dfrac{1}{100^{2}}<2$