\(a_n=\frac{1-\frac{1}{6}.\left(-\frac{n}{n+2}\right)^{n-3}}{1+\frac{1}{6}.\left(-\frac{n}{n+2}\right)^{n-3}}\)
\(a_n=\frac{1-\frac{1}{6}\left(\frac{1-n}{1+n}\right)^{n-3}}{1+\frac{1}{6}\left(\frac{1-n}{1+n}\right)^{n-3}}\)
cho n số thực dương \(a_{_{ }1},a_2,...,a_n\)có tổng bằng 1. Chứng minh rằng:
a) \(\left(a_1+\frac{1}{a_2}\right)^2+\left(a_2+\frac{1}{a_3}\right)^2+...+\left(a_n+\frac{1}{a_1}\right)^2\ge\left(\frac{n^2+1}{n}\right)^2\)
b) \(\left(a_1+\frac{1}{a_1}\right)^2+\left(a_2+\frac{1}{a_2}\right)^2+...+\left(a_n+\frac{1}{a_n}\right)^2\ge\left(\frac{n^2+1}{n}\right)^2\)
Cho dãy số \(\left(a_n\right)\) xác định bởi công thức:
\(\hept{\begin{cases}a_1=1;a_2=2;\\na_{n+2}=\left(3n+2\right)a_{n+1}-2\left(n+1\right)a_n;n=1;2;3...\end{cases}}\)
a) Tìm công thức số hạng tổng quát của dãy \(\left(a_n\right)\)
b)Chứng minh \(\sqrt{a_1-1}+\sqrt{a_2-1}+...+\sqrt{a_n-1}\ge\frac{n\left(n+1\right)}{2};\forall n\inℕ^∗\)
c) Tính \(lim\left(\frac{a_1}{3}+\frac{a_2}{3^2}+...+\frac{a_n}{3^n}\right)\)
\(a_n=\frac{1+\left(\frac{n}{n+2}\right)^n}{1-\left(\frac{n}{n+2}\right)^n}\)
\(a_n=\frac{1+\left(\frac{n}{n+2}\right)^n}{1-\left(\frac{n}{n+2}\right)^n}\)
\(a_n=\frac{\left(\frac{n}{n+2}\right)^2-\left(-1\right)}{\left(1-\frac{n}{n+2}\right)\left(1+\frac{n}{n+2}\right)}\)
\(a_n=\frac{\left(\frac{n}{n+2}-1\right)\left(\frac{n}{n+2}+1\right)}{\left(1-\frac{n}{n+2}\right)\left(1+\frac{n}{n+2}\right)}\)
\(a_n=\frac{\left(\frac{n}{n+2}-1\right)}{\left(1-\frac{n}{n+2}\right)}\)
\(a_n=\frac{-\left(1-\frac{n}{n+2}\right)}{\left(1-\frac{n}{n+2}\right)}\)
\(a_n=1\)
\(\Rightarrow\hept{\begin{cases}a=1\\n=1\end{cases}}\)
vậy \(\hept{\begin{cases}a=1\\n=1\end{cases}}\)
CMR với n thuộc N; n>=2 ta có:
\(A=\left(1-\frac{2}{6}\right)\left(1-\frac{2}{12}\right)\left(1-\frac{2}{20}\right)...\left(1-\frac{2}{n\left(n+1\right)}\right)>\frac{1}{3}\)\(\frac{1}{3}\)
với \(a_1,a_2,a_3,.....,a_n>0;a_1+a_2+a_3+....+a_n=k\)
Chứng minh\(\left(a_1+\frac{1}{a_2}\right)^2+\left(a_2+\frac{1}{a_3}\right)^2+...+\left(a_n+\frac{1}{a_1}\right)^2\ge\frac{1}{n}\left(\frac{k^2+n^2}{k}\right)^2\)
Chứng minh:\(\left(1-\frac{2}{6}\right)\left(1-\frac{2}{12}\right)\left(1-\frac{2}{20}\right)....\left(1-\frac{2}{n\left(n+1\right)}\right)>\frac{1}{3}\)
n thuộc n sao
hi !! ta cũng đang hỏi câu này -_-
chứng minh:
\(\left(1-\frac{2}{6}\right)\left(1-\frac{2}{12}\right)\left(1-\frac{2}{20}\right).....\left(1-\frac{2}{n\left(n+1\right)}\right)>\frac{1}{3\left(n\right)thuộcNsao}\)
\(1-\frac{2}{n\left(n+1\right)}=\frac{n^2+n-2}{n\left(n+1\right)}=\frac{\left(n-1\right)\left(n+2\right)}{n\left(n+1\right)}\)
\(\left(1-\frac{2}{2.3}\right)\left(...\right).....\left[1-\frac{2}{n\left(n+1\right)}\right]=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}.\frac{4.7}{5.6}....\frac{\left(n-2\right)\left(n+1\right)}{\left(n-1\right).n}.\frac{\left(n-1\right)\left(n+2\right)}{n\left(n+1\right)}=\)
\(=\frac{1}{3}.\frac{n+2}{n}=\frac{1}{3}-\frac{1}{3}.\frac{2}{n}>\frac{1}{3}\)
Cho a,b,c dương . CMR :
1) \(\frac{x^3}{y+z}+\frac{y^3}{x+z}+\frac{z^3}{x+y}\ge6;x+y+z\ge6\)
2) \(a_1.a_2....a_n\le\frac{1}{\left(n-1\right)^n};\frac{1}{a_1+1}+\frac{1}{a_2+1}+...+\frac{1}{a_n+1}=n-1\)
3) \(\frac{a}{b+c+1}+\frac{b}{a+c+1}+\frac{c}{b+a+1}+\left(1-a\right)\left(1-b\right)\left(1-c\right)\le1\) với a, b, c thuộc \(\left[0;1\right]\)