CMR: \(A=\dfrac{3}{4}+\dfrac{5}{36}+\dfrac{7}{144}+...+\dfrac{2n+1}{n^2\left(n+1\right)^2}\)< 1 ( n nguyên dương )
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
NT
26 tháng 9 2017
\(A=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+....+\dfrac{1}{18.19.20}=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{18.19}-\dfrac{1}{19.20}\right)\\ =\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{19.20}\right)\\ =\dfrac{1}{4}-\dfrac{1}{2.19.20}< \dfrac{1}{4}\)
Cái B TT nhé
\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+....+\dfrac{1}{n^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{\left(n-1\right)n}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n-1}-\dfrac{1}{n}\\ =1-\dfrac{1}{n}< 1\)
D TT
E mk thấy nó ss ớ
22 tháng 11 2018
a) Đặt \(A=\dfrac{1}{2^2}+\dfrac{1}{4^2}+\dfrac{1}{6^2}+...+\dfrac{1}{\left(2n\right)^2}\)
\(A=\dfrac{1}{2^2}\left(1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{n^2}\right)\)
Ta có:
\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{n^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{\left(n-1\right)n}\)
\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{n^2}< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n-1}-\dfrac{1}{n}\)
\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{n^2}< 1-\dfrac{1}{n}\)
\(\Rightarrow1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{n^2}< 1-\dfrac{1}{n}+1\)
\(\Rightarrow1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{n^2}< 2-\dfrac{1}{n}\)
\(\Rightarrow\dfrac{1}{2^2}\left(1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{n^2}\right)< \dfrac{1}{2^2}\left(2-\dfrac{1}{2}\right)\)
\(\Rightarrow A< \dfrac{1}{2^2}.2-\dfrac{1}{2^2}.\dfrac{1}{2}\)
\(\Rightarrow A< \dfrac{1}{2}-\dfrac{1}{2^3}< \dfrac{1}{2}\)
Vậy \(A< \dfrac{1}{2}\left(Đpcm\right)\)
b) Đặt \(B=\dfrac{1}{3^2}+\dfrac{1}{5^2}+\dfrac{1}{7^2}+...+\dfrac{1}{\left(2n+1\right)^2}\)
Ta có:
\(B< \dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{\left(2n-1\right)\left(2n+1\right)}\)
\(B< \dfrac{1}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{\left(2n-1\right)\left(2n+1\right)}\right)\)
\(B< \dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\right)\)
\(B< \dfrac{1}{2}\left(1-\dfrac{1}{2n+1}\right)\)
\(B< \dfrac{1}{2}\left(\dfrac{2n+1}{2n+1}-\dfrac{1}{2n+1}\right)\)
\(B< \dfrac{1}{2}.\dfrac{2n}{2n+1}\)
\(B< \dfrac{2n}{4n+2}\)
\(B< \dfrac{2n}{2\left(2n+1\right)}\)
\(B< \dfrac{n}{2n+1}\)
AH
Akai Haruma
Giáo viên
24 tháng 3 2018
Lời giải:
Ta có: \(4+(2n-1)^4=[(2n-1)^2+2]^2-[2(2n-1)]^2\)
\(=[(2n-1)^2+2-2(2n-1)][(2n-1)^2+2+2(2n-1)]\)
\(\Rightarrow \frac{2n-1}{4+(2n-1)^4}=\frac{2n-1}{[(2n-1)^2+2-2(2n-1)][(2n-1)^2+2+2(2n-1)]}\)
\(=\frac{1}{4}\left(\frac{1}{(2n-1)^2+2-2(2n-1)}-\frac{1}{(2n-1)^2+2+2(2n-1)}\right)\)
Do đó:
\(\frac{1}{4+1^4}=\frac{1}{4}(1-\frac{1}{5})\)
\(\frac{3}{4+3^4}=\frac{1}{4}(\frac{1}{5}-\frac{1}{17})\)
\(\frac{5}{4+5^4}=\frac{1}{4}(\frac{1}{17}-\frac{1}{37})\)
......
Do đó:
\(\frac{1}{4+1^4}+\frac{3}{4+3^4}+...+\frac{2n-1}{4+(2n-1)^4}=\frac{1}{4}(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{17}+...+\frac{1}{(2n-1)^2+2-2(2n-1)}-\frac{1}{(2n-1)^2+2+2(2n-1)})\)
\(=\frac{1}{4}(1-\frac{1}{(2n-1)^2+2+2(2n-1)})=\frac{1}{4}(1-\frac{1}{(2n-1+1)^2+1})\)
\(=\frac{1}{4}(1-\frac{1}{4n^2+1})=\frac{n^2}{4n^2+1}\)
Ta có đpcm.
25 tháng 3 2018
n=1 ; \(\dfrac{1}{4+1^4}=\dfrac{1}{5}=\dfrac{1^2}{4.^2+1}=\dfrac{1}{5};dung\)
giả sử n =k đúng \(\Leftrightarrow S=\dfrac{1}{4+1^4}+...+\dfrac{2k-1}{4+\left(2k-1\right)^4}=\dfrac{k^2}{4k^2+1}\) (*)
cần c/m đúng n =k+1 ;
c/m
với n=k+1
\(S=\left(\dfrac{1}{4+1^4}+...+\dfrac{2k-1}{4+\left(2k-1\right)^4}\right)+\dfrac{2\left(k+1\right)-1}{4+\left(2\left(k+1\right)-1\right)^4}\)
từ (*) =>\(S=\dfrac{k^2}{4k^2+1}+\dfrac{2\left(k+1\right)-1}{4+\left(2\left(k+1\right)-1\right)^4}\)
\(k+1=t\Leftrightarrow k=t-1\)
\(S=\dfrac{t^2-2t+1}{4\left(t^2-2t+1\right)+1}+\dfrac{2t-1}{4+\left(2t-1\right)^4}\)
\(S=\dfrac{t^2-2t+2}{4t^2-8t+5}+\dfrac{2t-1}{\left(4t^2+1\right)\left(4t^2-8t+5\right)}=\dfrac{\left(t^2-2t+1\right)\left(4t^2+1\right)+2t-1}{\left(4t^2+1\right)\left(4t^2-8t+5\right)}\)\(S=\dfrac{t^2\left(4t^2-8t+5\right)}{\left(4t^2+1\right)\left(4t^2-8t+5\right)}=\dfrac{t^2}{\left(4t^2+1\right)}=\dfrac{\left(k+1\right)^2}{4\left(k+1\right)^2+1}\)
Vậy tổng trên đúng với k +1
theo Quy nạp ta có dpcm
S
10 tháng 3 2019
\(\left(\frac{1}{1}+\frac{1}{3}+\frac{1}{5}+....+\frac{1}{2n-1}\right)-\left(\frac{1}{2}+\frac{1}{4}+....+\frac{1}{2n}\right)=\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{2n-1}+\frac{1}{2n}\right)-2\left(\frac{1}{2}+\frac{1}{4}+.....+\frac{1}{2n}\right)=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2n-1}+\frac{1}{2n}-\frac{1}{1}-\frac{1}{2}-....-\frac{1}{n}=\frac{1}{n+1}+\frac{1}{n+2}+....+\frac{1}{2n}\left(\text{đpcm}\right)\)
NV
Nguyễn Việt Lâm
Giáo viên
20 tháng 1 2021
\(a=\lim\sqrt{n^3}\sqrt{\dfrac{1}{n^3}+\dfrac{2}{n^2}-1}=\infty.\left(-1\right)=-\infty\)
\(b=\lim\left(\sqrt{n^2+2n+3}-n+n-\sqrt[3]{n^2+n^3}\right)\)
\(=\lim\dfrac{2n+3}{\sqrt{n^2+2n+3}+n}+\lim\dfrac{-n^2}{n^2+n\sqrt[3]{n^2+n^3}+\sqrt[3]{\left(n^2+n^3\right)^2}}\)
\(=\lim\dfrac{2+\dfrac{3}{n}}{\sqrt{1+\dfrac{2}{n}+\dfrac{3}{n^2}}+1}+\lim\dfrac{-1}{1+\sqrt[3]{\dfrac{1}{n}+1}+\sqrt[3]{\left(\dfrac{1}{n}+1\right)^2}}=\dfrac{2}{2}-\dfrac{1}{3}=\dfrac{2}{3}\)
\(c=\lim\dfrac{\left(\dfrac{2}{\sqrt{n}}+\dfrac{1}{n}\right)\left(\dfrac{1}{\sqrt{n}}+\dfrac{3}{n}\right)}{\left(1+\dfrac{1}{n}\right)\left(1+\dfrac{2}{n}\right)}=\dfrac{0.0}{1.1}=0\)
NV
Nguyễn Việt Lâm
Giáo viên
20 tháng 1 2021
\(d=\lim\dfrac{4-3\left(\dfrac{2}{4}\right)^n}{9.\left(\dfrac{3}{4}\right)^n+\left(\dfrac{2}{4}\right)^n}=\dfrac{4}{0}=+\infty\)
\(e=\lim\dfrac{7-25\left(\dfrac{5}{7}\right)^n+3.\left(\dfrac{1}{7}\right)^n}{12.\left(\dfrac{6}{7}\right)^n-\left(\dfrac{3}{7}\right)^n+3\left(\dfrac{1}{7}\right)^n}=\dfrac{7}{0}=+\infty\)
\(f=\lim\dfrac{n^4-4n^6}{n\left(\sqrt{n^4+1}+\sqrt{4n^6+1}\right)}=\lim\dfrac{\dfrac{1}{n^2}-6}{\sqrt{\dfrac{1}{n^6}+\dfrac{1}{n^{10}}}+\sqrt{\dfrac{4}{n^4}+\dfrac{1}{n^{10}}}}=\dfrac{-6}{0}=-\infty\)
MH
14 tháng 9 2021
c)\(7^{2n}+7^{2n+2}=2450\)
⇒\(7^{2n}+7^{2n}.7^2=2450\)
⇒\(7^{2n}.50=2450\)
⇒\(7^{2n}=49\)\(=7^2\)
⇒2n=2
⇒n=1
22 tháng 2 2022
a: \(M=\dfrac{6}{5}+\dfrac{3}{2}\left(\dfrac{2}{5\cdot7}+...+\dfrac{2}{97\cdot99}+\dfrac{2}{99\cdot101}\right)\)
\(=\dfrac{6}{5}+\dfrac{3}{2}\left(\dfrac{1}{5}-\dfrac{1}{101}\right)\)
\(=\dfrac{6}{5}+\dfrac{3}{10}-\dfrac{3}{202}=\dfrac{150}{101}\)
b: 
AH
Akai Haruma
Giáo viên
8 tháng 3 2021
Đề bị lỗi công thức rồi bạn. Bạn cần viết lại để được hỗ trợ tốt hơn.
