chứng minh 1/9=1/25+1/49+...........+1/(2n+1)^2 <1/4
Những câu hỏi liên quan
(1-4/9)(1-4/25)(1-4/49)(1-4/81).....(1-4/(2n+1)2)
Cho n thuộc N* chứng minh 2/9+2/25+...+2/(2n+1)2 <1/2
Cho n thuộc N*. Chứng minh: 2/9+2/25+...+2/(2n+1)^2<1/2
Giúp nha, mai mk phải nộp rồi.
cho A = 1/4 + 1/9 + 1/16 + 1/25 +1/36 + 1/49 + 1/64 + 1/81 . Chứng tỏ A > 2/5
A=1/22+1/32+...+1/92
Ta có:1/22>1/2.3,1/32>1/3.4,...,1/92>1/9.10
⇒A>1/2.3+1/3.4+...+1/9.10
A>1/2-1/3+1/3-1/4+...+1/9-1/10
A>1/2-1/10
A>2/5(đpcm)
Đúng 1
Bình luận (0)
Ta có: A = 1/4 + 1/9 + 1/16 + 1/25 +1/36 + 1/49 + 1/64 + 1/81
Vì 1/22>1/2.3,1/32>1/3.4,...,1/92>1/9.10
=>A>1/2.3+1/3.4+...+1/9.10
=>A>1/2-1/3+1/3-1/4+...+1/9-1/10
=>A>1/2-1/10
=>A>2/5
Đúng 0
Bình luận (0)
Giải:
\(A=\dfrac{1}{4}+\dfrac{1}{9}+\dfrac{1}{16}+\dfrac{1}{25}+\dfrac{1}{36}+\dfrac{1}{49}+\dfrac{1}{64}+\dfrac{1}{81}\)
\(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2} +\dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+\dfrac{1}{8^2}+\dfrac{1}{9^2}\)
Ta có:
\(\dfrac{1}{2^2}=\dfrac{1}{2.2}>\dfrac{1}{2.3}\)
\(\dfrac{1}{3^2}=\dfrac{1}{3.3}>\dfrac{1}{3.4}\)
\(\dfrac{1}{4^2}=\dfrac{1}{4.4}>\dfrac{1}{4.5}\)
\(\dfrac{1}{5^2}=\dfrac{1}{5.5}>\dfrac{1}{5.6}\)
\(\dfrac{1}{6^2}=\dfrac{1}{6.6}>\dfrac{1}{6.7}\)
\(\dfrac{1}{7^2}=\dfrac{1}{7.7}>\dfrac{1}{7.8}\)
\(\dfrac{1}{8^2}=\dfrac{1}{8.8}>\dfrac{1}{8.9}\)
\(\dfrac{1}{9^2}=\dfrac{1}{9.9}>\dfrac{1}{9.10}\)
\(\Rightarrow A>\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}\)
\(\Rightarrow A>\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\)
\(\Rightarrow A>\dfrac{1}{2}-\dfrac{1}{10}\)
\(\Rightarrow A>\dfrac{2}{5}\left(đpcm\right)\)
Chúc bạn học tốt!
Đúng 0
Bình luận (0)
chứng minh rằng 2^2n(2^2n+1 -1) -1 chia hết cho 9
chứng minh 3^2n+1 +2^4n+1 chia hết cho 25
chứng minh rằng 2^2n * (2^2n+1 -1)-1 chia hết cho 9 với n thuộc N
minh van chua ro phan de 2^2n+1-1 la (2^2n+1) hay nhu de ghi ban a
Chứng minh rằng 22n(22n+1-1)-1 chia hết cho 9 với mọi n>1
22n(22n+1-1)-1
\(=2^{4n+1}-2^{2n}-1=2.2^{4n}-2^{2n}-1\)
\(=2\left(2^{2n}\right)^2-2^{2n}-1=A\)
Đặt \(2^{2n}=t\)
\(\Rightarrow A=2t^2-t-1=\left(2t+1\right)\left(t-1\right)\)
\(=\left(2.2^{2n}+1\right)\left(2^{2n}-1\right)\)
\(=\left(2^{2n+1}+1\right)\left(2^{2n}-1\right)=\left(2+1\right)\left(2^{2n}-2^{2n-1}+...+1\right)\left(2+1\right)\left(2^{2n-1}+...-1\right)\)
\(=9.B\)
Vậy \(A⋮9\)
Đúng 0
Bình luận (0)
chứng minh với mọi n\(\in\)N* và n\(\ge3\) có:
\(\dfrac{1}{9}+\dfrac{1}{25}+...+\dfrac{1}{\left(2n+1\right)^2}< \dfrac{1}{4}\)
Lời giải:
Ta thấy \((2n+1)^2=4n^2+4n+1> 4n^2+4n\)
\(\Leftrightarrow (2n+1)^2> 2n(2n+2)\) \(\Leftrightarrow \frac{1}{(2n+1)^2}\leq \frac{1}{2n(2n+2)}\)
Do đó:
\(\left\{\begin{matrix} \frac{1}{3^2}< \frac{1}{2.4}\\ \frac{1}{5^2}< \frac{1}{4.6}\\ .......\\ \frac{1}{(2n+1)^2}< \frac{1}{2n(2n+2)}\end{matrix}\right.\)
\(\Rightarrow \frac{1}{9}+\frac{1}{25}+....+\frac{1}{(2n+1)^2}< \frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{2n(2n+2)}=M\) (1)
\(2M=\frac{2}{2.4}+\frac{2}{4.6}+....+\frac{2}{2n(2n+2)}\)
\(=\frac{4-2}{2.4}+\frac{6-4}{4.6}+\frac{8-6}{6.8}+....+\frac{2n+2-2n}{2n(2n+2)}\)
\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-...+\frac{1}{2n}-\frac{1}{2n+2}\)
\(=\frac{1}{2}-\frac{1}{2n+2}< \frac{1}{2}\)
\(\Rightarrow M< \frac{1}{4} (2)\)
Từ (1),(2) suy ra \(\frac{1}{9}+\frac{1}{25}+...+\frac{1}{(2n+1)^2}< \frac{1}{4}\) (đpcm)
Đúng 0
Bình luận (0)