Cho S=1/2^2+1/3^2+1/4^2+…+1/2015^2.Chưng tỏ rằng 1007/2016 { "@context": "https://schema.org", "@type": "QAPage", "mainEntity": { "@type": "Question", "name": " Cho S=1/2^2+1/3^2+1/4^2+...+1/2015^2.Chưng tỏ rằng 1007/2016 " 5<s<8="" cho="" mk="" ctlhn="" ạ="" giải="" thích="" các="" bước="" giải:",="" 9:06:57="" "url":="" "https:="" mtrend.vn="" cho-s-1-2-2-1-3-2-1-4-2-1-2015-2-chung-to-rang-1007-2016-s-2014-2015-272="" #comment-17760",="" "author":="" "person",="" "url"="" :="" author="" hiennhi",="" "name":="" "hiennhi"="" }="" },{="" "ta="" có="" frac{1}{2^2}$="" $="" frac{1}{3^2}$="" frac{1}{4^2}$="" frac{1}{2015^2}$="" >="" frac{1}{2.3}$="" frac{1}{3.4}$="" frac{1}{4.5}$="" frac{1}{2015.2016}$="" ↔="" frac{3-2}{2.3}$="" frac{4-3}{3.4}$="" frac{5-4}{4.5}$="" frac{2016-2015}{2015.2016}$="" frac{1}{2}$="" frac{1}{3}$="" frac{1}{4}$="" frac{1}{5}$="" frac{1}{2015}$="" frac{1}{2016}$="" frac{1007}{2016}$="" ------------------------------------------------------------------------------------------="" <="" frac{1}{1.2}$="" frac{2014}{2015}$="" frac{2-1}{1.2}$="" frac{2015-2014}{2014.2015}$="" frac{1}{2014}$="" vậy="" ta="" đpcm.",="" 9:07:24="" #comment-17761",="" anhthu",="" "anhthu"="" ]="" <="" script="">
Đáp án:
S<1/2^2 + 1/2.3 + 1/3.4 +…+ 1/8.9
S<1/4 + 1/2 – 1/3 + 1/3 – 1/4+…+1/8 – 1/9
S<1/4 + 1/2 – 1/9
S<23/36<8/9 (1)
Mặt khác: S>1/2^2 + 1/3.4 + …+ 1/9*10
S>1/4 + 1/3 – 1/4 + … + 1/9 – 1/10
S>1/4 + 1/3 – 1/10
S>29/60>2/5 (2)
Từ (1),(2)
=> 2/5<S<8/9
cho mk ctlhn ạ
Giải thích các bước giải:
ta có
S = $\frac{1}{2^2}$ + $\frac{1}{3^2}$ + $\frac{1}{4^2}$ + … + $\frac{1}{2015^2}$
S > $\frac{1}{2.3}$ + $\frac{1}{3.4}$ + $\frac{1}{4.5}$ + … + $\frac{1}{2015.2016}$
⇔ S > $\frac{3-2}{2.3}$ + $\frac{4-3}{3.4}$ + $\frac{5-4}{4.5}$ + … + $\frac{2016-2015}{2015.2016}$
⇔ S > $\frac{1}{2}$ – $\frac{1}{3}$ + $\frac{1}{3}$ – $\frac{1}{4}$ + $\frac{1}{4}$ – $\frac{1}{5}$ + … + $\frac{1}{2015}$ – $\frac{1}{2016}$
⇔ S > $\frac{1}{2}$ – $\frac{1}{2016}$ = $\frac{1007}{2016}$
——————————————————————————————
S = $\frac{1}{2^2}$ + $\frac{1}{3^2}$ + $\frac{1}{4^2}$ + … + $\frac{1}{2015^2}$
S < $\frac{1}{1.2}$ + $\frac{1}{2.3}$ + $\frac{1}{3.4}$ + … + $\frac{2014}{2015}$
⇔ S < $\frac{2-1}{1.2}$ + $\frac{3-2}{2.3}$ + $\frac{4-3}{3.4}$ + … + $\frac{2015-2014}{2014.2015}$
⇔ S < 1 – $\frac{1}{2}$ + $\frac{1}{2}$ – $\frac{1}{3}$ + $\frac{1}{3}$ – … + $\frac{1}{2014}$ – $\frac{1}{2015}$
⇔ S < 1 – $\frac{1}{2015}$ = $\frac{2014}{2015}$
Vậy ta có đpcm.