chứng tỏ rằng:1/5 + 1/13 +1/25 +...+1/n2+(n+1)2 < 1/2 với mọi n thuộc N
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NN
1
21 tháng 7 2016
\(A=\frac{1}{5}+\frac{1}{13}+\frac{1}{25}+...+\frac{1}{2.n^2+2n+1}< \frac{1}{4}+\frac{1}{12}+\frac{1}{24}+...+\frac{1}{2.n^2+2n}\)
\(A< \frac{1}{2}.\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{n.\left(n+1\right)}\right)\)
\(A< \frac{1}{2}.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{n.\left(n+1\right)}\right)\)
\(A< \frac{1}{2}.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-...+\frac{1}{n}-\frac{1}{n+1}\right)\)
\(A< \frac{1}{2}.\left(1-\frac{1}{n+1}\right)< \frac{1}{2}\)
\(\Rightarrow A< \frac{1}{2}\)
18 tháng 7 2016
\(A=\frac{1}{5}+\frac{1}{13}+\frac{1}{25}+...+\frac{1}{2.n^2+2n+1}< \frac{1}{4}+\frac{1}{12}+\frac{1}{24}+...+\frac{1}{2.n^2+2n}\)
\(A< \frac{1}{2}.\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{n.\left(n+1\right)}\right)\)
\(A< \frac{1}{2}.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{n.\left(n+1\right)}\right)\)
\(A< \frac{1}{2}.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n}-\frac{1}{n+1}\right)\)
\(A< \frac{1}{2}.\left(1-\frac{1}{n+1}\right)< \frac{1}{2}\)
=> \(A< \frac{1}{2}\)
CC
21 tháng 7 2019
1, Thấy : \(\frac{1}{5}< \frac{2}{2.4}\)
\(\frac{1}{13}< \frac{2}{4.6}\)
.....
\(\frac{1}{n^2+\left(n+1\right)^2}< \frac{2}{2n\left(2n+1\right)}\)
Cộng từng vế có :
\(\frac{1}{5}+\frac{1}{13}+...+\frac{1}{n^2+\left(n+1\right)^2}< \frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{2n\left(2n+2\right)}\)
\(\frac{1}{5}+\frac{1}{13}+...+\frac{1}{n^2+\left(n+1\right)^2}< \frac{1}{2}-\frac{1}{4}+....+\frac{1}{2n}-\frac{1}{2n+2}\)
\(\frac{1}{5}+\frac{1}{13}+..+\frac{1}{n^2+\left(n+1\right)^2}< \frac{1}{2}-\frac{1}{2n+2}\)
Mà \(\frac{1}{2}-\frac{1}{2n+2}< \frac{1}{2}\)=> Tổng trên < 1/2
CC
21 tháng 7 2019
2,M = \(\frac{3}{\left(1.2\right)^2}+\frac{5}{\left(2.3\right)^2}+...+\frac{2n+1}{\left[n\left(n+1\right)\right]^2}\)
=> M \(=\frac{1}{1^2}-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+...+\frac{1}{\left(n-1\right)^2}-\frac{1}{n^2}+\frac{1}{n^2}-\frac{1}{\left(n+1\right)^2}\)
\(M=1-\frac{1}{\left(n+1\right)^2}=\frac{\left(n+1\right)^2-1}{\left(n+1\right)^2}=\frac{n^2+2n+1-1}{\left(n+1\right)^2}=\frac{n^2+2n}{\left(n+1\right)^2}\)
Đến đây tắc r tự nghĩ tiếp >:
TQ
1
24 tháng 4 2016
M = 1/2.2 + 1/3.3 +.....+ 1/n.n
M < 1/1.2 + 1/2.3 +.....+ 1/(n-1).n
M < 1 - 1/2 + 1/2 - 1/3 +......+ 1/n-1 - 1/n
M < 1 - 1/n < 1
=> M < 1 (đpcm)
Ai k mk mk k lại cho,kết bạn luôn nhé!
PH
0
NT
0
Ta có: \(n^2+\left(n+1\right)^2>2n\left(n+1\right)\)
\(\Rightarrow\frac{1}{5}+\frac{1}{13}+...+\frac{1}{n^2+\left(n+1\right)^2}\)
\(=\frac{1}{1^2+2^2}+\frac{1}{2^2+3^2}+...+\frac{1}{n^2+\left(n+1\right)^2}< \frac{1}{2.1.2}+\frac{1}{2.2.3}+...+\frac{1}{2.n.\left(n+1\right)}\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{n.\left(n+1\right)}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{n+1}\right)< \frac{1}{2}\)