chứng minh rằng 1/3 mu 2 +1/4 mu 2 +1/5 mu 2+ 1/6 mu 2+….1/100 mu 2<1/2 28/09/2021 Bởi Ivy chứng minh rằng 1/3 mu 2 +1/4 mu 2 +1/5 mu 2+ 1/6 mu 2+….1/100 mu 2<1/2
`1/3^2+1/4^2+1/5^2+…+1/100^2 < 1/(2.3)+1/(3.4)+1/(4.5)+…+1/(99.100)` `1/3^2+1/4^2+1/5^2+…+1/100^2 <1/2-1/3+1/3-1/4+1/4-1/5+..+1/99-1/100` `1/3^2+1/4^2+1/5^2+…+1/100^2 <1/2-1/100 < 1/2` Bình luận
Đáp án: Giải thích các bước giải: Ta có: $\dfrac{1}{3^{2}}<\dfrac{1}{2.3}$ $;$ $\dfrac{1}{4^{2}}<\dfrac{1}{3.4}$ $;…;$ $\dfrac{1}{100^{2}}<\dfrac{1}{99.100}$ $ $ $⇒\dfrac{1}{3^{2}}+\dfrac{1}{4^{2}}+…+\dfrac{1}{100^{2}}<\dfrac{1}{2.3}+\dfrac{1}{3.4}+…+\dfrac{1}{99.100}$ $ $ $⇒\dfrac{1}{3^{2}}+\dfrac{1}{4^{2}}+…+\dfrac{1}{100^{2}}<\dfrac{1}{2}-\dfrac{1}{100}<\dfrac{1}{2}$ $ $ $⇒\dfrac{1}{3^{2}}+\dfrac{1}{4^{2}}+…+\dfrac{1}{100^{2}}<\dfrac{1}{2}$ Bình luận
`1/3^2+1/4^2+1/5^2+…+1/100^2 < 1/(2.3)+1/(3.4)+1/(4.5)+…+1/(99.100)`
`1/3^2+1/4^2+1/5^2+…+1/100^2 <1/2-1/3+1/3-1/4+1/4-1/5+..+1/99-1/100`
`1/3^2+1/4^2+1/5^2+…+1/100^2 <1/2-1/100 < 1/2`
Đáp án:
Giải thích các bước giải:
Ta có: $\dfrac{1}{3^{2}}<\dfrac{1}{2.3}$ $;$ $\dfrac{1}{4^{2}}<\dfrac{1}{3.4}$ $;…;$ $\dfrac{1}{100^{2}}<\dfrac{1}{99.100}$
$ $
$⇒\dfrac{1}{3^{2}}+\dfrac{1}{4^{2}}+…+\dfrac{1}{100^{2}}<\dfrac{1}{2.3}+\dfrac{1}{3.4}+…+\dfrac{1}{99.100}$
$ $
$⇒\dfrac{1}{3^{2}}+\dfrac{1}{4^{2}}+…+\dfrac{1}{100^{2}}<\dfrac{1}{2}-\dfrac{1}{100}<\dfrac{1}{2}$
$ $
$⇒\dfrac{1}{3^{2}}+\dfrac{1}{4^{2}}+…+\dfrac{1}{100^{2}}<\dfrac{1}{2}$