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
Cho S=1/2^2+1/3^2+1/4^2+…+1/2015^2.Chưng tỏ rằng 1007/2016
By Allison
By Allison
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.