Cho \(a_n=\left(-1\right)^n\cdot\left(\frac{n^2+n+1}{n\text{!}}\right)\)Với n > 0.
Tính \(A=a_1+a_2+a_3+...+a_{2017}\)
CMR:
Nếu \(\frac{a_1}{a_2}=\frac{a_2}{a_3}=\frac{a_3}{a_4}=...=\frac{a_n}{a_{n+1}}\)thì\(\left(\frac{a_1+a_2+a_3+...+a_n}{a_2+a_3+a_4+..+a_{n+1}}\right)^n=\frac{a_1}{a_{n+1}}\)
áp dụng t.c dãy tỉ số bằng nhau ta có:
\(\frac{a1}{a2}=\frac{a2}{a3}=\frac{a3}{a4}=.....=\frac{an}{an+1}=\frac{a1+a2+a3+....+an}{a2+a3+a4+...+an+1}\)
\(\frac{a1}{a2}\cdot\frac{a2}{a3}\cdot\frac{a3}{a4}\cdot...\cdot\frac{an}{an+1}=\frac{a1}{an+1}=\left(\frac{a1}{a2}\right)^n=\left(\frac{a1+a2+a3+....+an}{a2+a3+a4+...+an+1}\right)^n\)(vì từ 1 đến n có n chữ số)
=> đpcm
(Nghi binh 20/09)
Cho \(a_1,a_2,...,a_n>0;3\le n\in N.\) Đặt:
\(A_1=\frac{a_1}{a_2+a_3}+\frac{a_2}{a_3+a_4}+...+\frac{a_{n-1}}{a_n+a_1}+\frac{a_n}{a_1+a_2}\)
\(A_2=\frac{a_1}{a_n+a_2}+\frac{a_2}{a_1+a_3}+...+\frac{a_{n-1}}{a_{n-2}+a_n}+\frac{a_n}{a_{n-1}+a_1}\)
Chứng minh rằng: \(Max\left\{A_1,A_2\right\}\ge\frac{n}{2}\)
Cho \(\hept{\begin{cases}a_1>a_2>...>a_n>0\\1\le k\in Z\end{cases}}\)
CMR : \(a_1+\frac{1}{a_n\left(a_1-a_2\right)^k\left(a_2-a_3\right)^k...\left(a_{n-1}-a_n\right)^k}\ge\frac{\left(n-1\right)k+2}{\sqrt[\left(n-1\right)k+2]{k^{\left(n-1\right)k}}}\)
với \(a_1,a_2,a_3,.....,a_n>0;a_1+a_2+a_3+....+a_n=k\)
Chứng minh\(\left(a_1+\frac{1}{a_2}\right)^2+\left(a_2+\frac{1}{a_3}\right)^2+...+\left(a_n+\frac{1}{a_1}\right)^2\ge\frac{1}{n}\left(\frac{k^2+n^2}{k}\right)^2\)
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: \(\frac{a_1}{a_2}=\frac{a_2}{a_3}=\frac{a_3}{a_4}=...=\frac{a_{n-1}}{a_n}=\frac{a_n}{a_1}\) với \(a_1+a_2+...+a_n\)# 0. Tính:
1. A = \(\frac{a^2_1+a^2_2+...+a^2_n}{\left(a_1+a_2+...+a_n\right)^2}\)
2. B = \(\frac{a^9_1+a^9_2+...+a^9_n}{\left(a_1+a_2+...+a_n\right)^9}\)
cho \(\frac{a_1}{a_2}=\frac{a_2}{a_3}=...=\frac{a_{n-1}}{a_n}=\frac{a_n}{a_1};a_1+a_2+..+a_{n-1}+a_n\ne0\)
Tính \(\frac{a^2_2+a^2_2+...+a^2_n}{\left(a_1+a_2+...+a_n\right)^2}\)
Đặt \(\frac{a_1}{a_2}=\frac{a_2}{a_3}=...=\frac{a_{n-1}}{a_n}=\frac{a_n}{a_1}=k\)
=>\(\frac{a_1}{a_2}.\frac{a_2}{a_3}.....\frac{a_{n-1}}{a_n}.\frac{a_n}{a_1}=k.k.....k.k\)
=>\(k^n=\frac{a_1.a_2.....a_{n-1}.a_n}{a_2.a_3.....a_n.a_1}\)
=>\(k^n=1=1^n\)
=>k=1
=>\(\frac{a_1}{a_2}=\frac{a_2}{a_3}=...=\frac{a_{n-1}}{a_n}=\frac{a_n}{a_1}=1\)
=>\(a_1=a_2=...=a_n\)
\(=>\frac{a^2_1+a^2_2+...+a_n^2}{\left(a_1+a_2+...+a_n\right)^2}\)
=\(\frac{a^2_1+a^2_1+...+a_1^2}{\left(a_1+a_1+...+a_1\right)^2}\)
=\(\frac{n.a^2_1}{\left(n.a_1\right)^2}=\frac{n.a_1^2}{n^2.a^2_1}=\frac{1}{n}\)
thế này dc ko
Áp dụng t/c của dãy tỉ số bằng nhau, ta có :
\(\frac{a_1}{a_2}=\frac{a_2}{a_3}=...=\frac{a_{n-1}}{a_n}=\frac{a_n}{a_1}=\frac{a_1+a_2+...+a_{n-1}+a_n}{a_2+a_3+...+a_n+a_1}\Rightarrow a_1=a_2=...=a_n\)
\(\frac{a^1_2+a^2_2+...+a^2_n}{\left(a_1+a_2+...+a_n\right)}=\frac{na^2_1}{\left(na_1\right)^2}=\frac{1}{n}\)
Chứng minh rằng nếu: \(\frac{a_1}{a_2}\)=\(\frac{a_2}{a_3}\)=...=\(\frac{a_n}{a_{n+1}}\)(n \(\varepsilon\)\(ℕ^∗\))
thì:\(\frac{\left(a_1+a_2+...+a_n\right)^n}{\left(a_2+a_3+...+a_{n+1}\right)^n}\)=\(\frac{a^n_1+a^n_2+...+a_n^n}{a^n_2+...+a_{n+1}^n}\)=\(\frac{a_1}{a_{n+1}}\)
Ta có :
\(\frac{a_1}{a_2}=\frac{a_2}{a_3}=...=\frac{a_n}{a_{n+1}}=\frac{a_1+a_2+...+a_n}{a_2+a_3+...+a_{n+1}}\)
\(\Rightarrow\)\(\frac{a_1^n}{a_2^n}=\frac{a_2^n}{a_3^n}=...=\frac{a_n^n}{a_{n+1}^n}=\frac{a_1^n+a_2^n+...+a_n^n}{a_2^n+a_3^n+...+a_{n+1}^n}=\frac{\left(a_1+a_2+...+a_n\right)^n}{\left(a_2+a_3+...+a_{n+1}\right)^n}=\frac{a_1.a_2...a_n}{a_2.a_3...a_{n+1}}=\frac{a_1}{a_{n+1}}\)
Chứng minh rằng nếu \(\dfrac{a_1}{a_2}=\dfrac{a_2}{a_3}=...=\dfrac{a_n}{a_{n+1}}\) thì \(\left(\dfrac{a_1+a_2+a_3+...+a_n}{a_2+a_3+a_4+...+a_{n+1}}\right)^n=\dfrac{a_1}{a_{n+1}}\)
Theo tính chất của dãy tỉ số bằng nha, ta có :
\(\dfrac{a_1}{a_2}=\dfrac{a_2}{a_3}=.....=\dfrac{a_n}{a_{n+1}}=\dfrac{a_1+a_2+....+a_n}{a_2+a_3+....+a_{n+1}}\)
\(\Rightarrow\dfrac{a_1}{a_2}=\dfrac{a_1+a_2+....+a_n}{a_2+a_3+....+a_{n+1}}\)
\(\dfrac{a_2}{a_3}=\dfrac{a_1+a_2+.....+a_n}{a_2+a_3+.....+a_{n+1}}\)
.................................
\(\dfrac{a_n}{a_{n+1}}=\dfrac{a_1+a_2+.....+a_n}{a_2+a_3+.....+a_{n+1}}\)
\(\Rightarrow\left(\dfrac{a_1+a_2+.....+a_n}{a_2+a_3+.....+a_{n+1}}\right)^n=\dfrac{a_1}{a_2}.\dfrac{a_2}{a_3}........\dfrac{a_n}{a_{n+1}}\)
Vậy \(\left(\dfrac{a_1+a_2+......+a_n}{a_2+a_3+......+a_{n+1}}\right)=\dfrac{a_1}{a_{n+1}}\) (đpcm)
~ Học tốt ~