K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

AH
Akai Haruma
Giáo viên
23 tháng 11 2017

Lời giải:

Sử dụng quy nạp:

Với \(n=1\Rightarrow \frac{1}{2}< \frac{1}{\sqrt{3}}\) (đúng)

Với \(n=2\Rightarrow \frac{1.3}{2.4}< \frac{1}{\sqrt{5}}\) (đúng)

.............

Giả sử bài toán đúng với \(n=k\), tức là :

\(\frac{1.3.5...(2k-1)}{2.4.6...2k}< \frac{1}{\sqrt{2k+1}}\) (*)

Ta cần chỉ ra nó cũng đúng với \(n=k+1\) hay :

\(\frac{1.3.5....(2k-1)(2k+1)}{2.4.6....(2k)(2k+2)}< \frac{1}{\sqrt{2k+3}}\). Thật vậy, theo (*) ta có:

\(\frac{1.3.5....(2k-1)(2k+1)}{2.4.6....(2k)(2k+2)}< \frac{1}{\sqrt{2k+1}}.\frac{2k+1}{2k+2}=\frac{\sqrt{2k+1}}{2k+2}\) (1)

Xét \(\frac{\sqrt{2k+1}}{2k+2}-\frac{1}{\sqrt{2k+3}}=\frac{\sqrt{(2k+1)(2k+3)}-(2k+2)}{(2k+2)\sqrt{2k+3}}\) \(=\frac{-1}{[\sqrt{(2k+1)(2k+3)}+(2k+2)](2k+2)\sqrt{2k+3}}<0\)

Suy ra \(\frac{\sqrt{2k+1}}{2k+2}< \frac{1}{\sqrt{2k+3}}(2)\)

Từ \((1);(2)\Rightarrow \frac{1.3.5....(2k-1)(2k+1)}{2.4.6....(2k)(2k+2)}< \frac{1}{\sqrt{2k+3}}\)

Vậy bài toán đúng với \(n=k+1\), phép quy nạp hoàn thành.

Do đó ta có đpcm.

NV
27 tháng 10 2019

Làm biếng gõ lại:

Câu hỏi của dbrby - Toán lớp 10 | Học trực tuyến

NV
4 tháng 12 2021

\(A_n=\dfrac{\sqrt{2n-1}}{\left(2n+1\right)\left(2n-1\right)}=\dfrac{\sqrt{2n-1}}{2}\left(\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\right)\)

\(=\dfrac{\sqrt{2n-1}}{2}\left(\dfrac{1}{\sqrt{2n-1}}-\dfrac{1}{\sqrt{2n+1}}\right)\left(\dfrac{1}{\sqrt{2n-1}}+\dfrac{1}{\sqrt{2n+1}}\right)\)

\(< \dfrac{\sqrt{2n-1}}{2}\left(\dfrac{1}{\sqrt{2n-1}}-\dfrac{1}{\sqrt{2n+1}}\right)\left(\dfrac{1}{\sqrt{2n-1}}+\dfrac{1}{\sqrt{2n-1}}\right)\)

\(=\dfrac{1}{\sqrt{2n-1}}-\dfrac{1}{\sqrt{2n+1}}\)

\(\Rightarrow A_1+A_2+...+A_n< 1-\dfrac{1}{\sqrt{3}}+\dfrac{1}{\sqrt{3}}-\dfrac{1}{\sqrt{5}}+...+\dfrac{1}{\sqrt{2n-1}}-\dfrac{1}{\sqrt{2n+1}}=1-\dfrac{1}{\sqrt{2n+1}}< 1\)

NA
Ngoc Anh Thai
Giáo viên
28 tháng 3 2021

a) Vế trái  \(=\dfrac{1.3.5...39}{21.22.23...40}=\dfrac{1.3.5.7...21.23...39}{21.22.23....40}=\dfrac{1.3.5.7...19}{22.24.26...40}\)

               \(=\dfrac{1.3.5.7....19}{2.11.2.12.2.13.2.14.2.15.2.16.2.17.2.18.2.19.2.20}\\ =\dfrac{1.3.5.7.9.....19}{\left(1.3.5.7.9...19\right).2^{20}}=\dfrac{1}{2^{20}}\left(đpcm\right)\)

b) Vế trái

 \(=\dfrac{1.3.5...\left(2n-1\right)}{\left(n+1\right).\left(n+2\right).\left(n+3\right)...2n}\\ =\dfrac{1.2.3.4.5.6...\left(2n-1\right).2n}{2.4.6...2n.\left(n+1\right)\left(n+2\right)...2n}\\ =\dfrac{1.2.3.4...\left(2n-1\right).2n}{2^n.1.2.3.4...n.\left(n+1\right)\left(n+2\right)...2n}\\ =\dfrac{1}{2^n}.\\ \left(đpcm\right)\)

              

23 tháng 8 2016

có ai giúp với

31 tháng 7 2018

Ta có: \(\dfrac{1}{9}=\left(\dfrac{1}{3}\right)^2=\dfrac{1}{3.3}< \dfrac{1}{2.3}\)

\(\dfrac{1}{16}=\left(\dfrac{1}{4}\right)^2=\dfrac{1}{4.4}< \dfrac{1}{3.4}\)

................

\(\dfrac{1}{\left(2n+1\right)^2}< \dfrac{1}{2n\left(2n+1\right)}\)

\(\dfrac{1}{9}+\dfrac{1}{16}+......+\dfrac{1}{\left(2n+1\right)^2}\)< \(\dfrac{1}{2.3}+\dfrac{1}{3.4}+.....+\dfrac{1}{2n.\left(2n+1\right)}\)

= \(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+.....+\dfrac{1}{2n}-\dfrac{1}{2n+1}\)

= \(\dfrac{1}{2}-\dfrac{1}{2n+1}\)

= \(\dfrac{2n+1-2}{2n+1}\)

= \(\dfrac{2n-1}{2n+1}\)= \(1-\dfrac{2}{2n+1}\)

Ta có: n ≥ 1⇒ 2n+1 ≥ 3

\(1-\dfrac{2}{2n+1}\)\(\dfrac{1}{3}\)

hình như đề sai thì phải

NV
22 tháng 1 2019

\(\dfrac{1}{\left(3n-1\right)\left(3n+2\right)}=\dfrac{1}{3}\left(\dfrac{1}{3n-1}-\dfrac{1}{3n+2}\right)\)

\(\Rightarrow A=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{3n-1}-\dfrac{1}{3n+2}\right)\)

\(\Rightarrow A=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{3n+2}\right)\)

\(\Rightarrow A=\dfrac{3n}{6\left(3n+2\right)}=\dfrac{n}{6n+4}\)

\(\dfrac{1}{\left(2n-1\right)\left(2n+1\right)\left(2n+3\right)}=\dfrac{1}{4}\left(\dfrac{1}{\left(2n-1\right)\left(2n+1\right)}-\dfrac{1}{\left(2n+1\right)\left(2n+3\right)}\right)\)

\(\Rightarrow B=\dfrac{1}{4}\left(\dfrac{1}{1.3}-\dfrac{1}{3.5}+\dfrac{1}{3.5}-\dfrac{1}{3.7}+...+\dfrac{1}{\left(2n-1\right)\left(2n+1\right)}-\dfrac{1}{\left(2n+1\right)\left(2n+3\right)}\right)\)

\(\Rightarrow B=\dfrac{1}{4}\left(\dfrac{1}{1.3}-\dfrac{1}{\left(2n+1\right)\left(2n+3\right)}\right)\)

\(\Rightarrow B=\dfrac{n\left(n+2\right)}{3\left(2n+1\right)\left(2n+3\right)}\)

\(\sqrt{1+\dfrac{1}{n^2}+\dfrac{1}{\left(n+1\right)^2}}=\sqrt{\dfrac{n^2\left(n+1\right)^2+\left(n+1\right)^2+n^2}{n^2\left(n+1\right)^2}}\)

\(=\sqrt{\dfrac{n^2\left(n+1\right)^2+2n^2+2n+1}{n^2\left(n+1\right)^2}}=\sqrt{\dfrac{n^2\left(n+1\right)^2+2n\left(n+1\right)+1}{n^2\left(n+1\right)^2}}\)

\(=\sqrt{\dfrac{\left[n\left(n+1\right)+1\right]^2}{n^2\left(n+1\right)^2}}=\dfrac{n\left(n+1\right)+1}{n\left(n+1\right)}=1+\dfrac{1}{n\left(n+1\right)}=1+\dfrac{1}{n}-\dfrac{1}{n+1}\)

\(\Rightarrow C=1+\dfrac{1}{1}-\dfrac{1}{2}+1+\dfrac{1}{2}-\dfrac{1}{3}+1+\dfrac{1}{3}-\dfrac{1}{4}+...+1+\dfrac{1}{2018}-\dfrac{1}{2019}\)

\(\Rightarrow C=2019-\dfrac{1}{2019}\)

22 tháng 1 2019

@Luân Đào @Nguyễn Việt Lâm