Với \(a\in Z^+\)đặt \(a_n=\frac{4n}{n^4+4}\)
Chứng minh: \(a_1+a_2+a_3+...+a_{2015}<\frac{3}{4}\)
(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: \(Max\left\{A_1,A_2\right\}\ge\frac{n}{2}\)
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
Cho n số khác 0 là a1, a2, a3,....,an thảo mãn \(a_2^2=a_1.a_3,a_3^2=a_2.a_4,...,a_{n-1}^2=a_{n-2}.a_n\). Chứng minh \(\frac{a_1^3+a_2^3+a_3^3+...+a_{n-1}^3}{a_2^3+a_3^3+a_4^3+...+a_n^3}=\frac{a_1}{a_n}\)
Tính tổng A =\(\frac{c}{a_1.a_2}+\frac{c}{a_2.a_3}+....+\frac{c}{a_{n-1}.a_n}\)với \(a_2-a_1=a_3-a_2=...=a_n-a_{n-1}=k\)
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 ~
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 :
\(\frac{a_1}{a_2}=\frac{a_2}{a_3}=...=\frac{a_{n-1}}{a_n}=\frac{a_n}{a_1}=\frac{a_1+a_2+...+a_{n-1}+a_n}{a_2+a_3+...+a_n+a_1}\Rightarrow a_1=a_2=...=a_n\)
\(\frac{a^1_2+a^2_2+...+a^2_n}{\left(a_1+a_2+...+a_n\right)}=\frac{na^2_1}{\left(na_1\right)^2}=\frac{1}{n}\)
\(Cho\) \(\dfrac{a_1}{a_2}=\dfrac{a_2}{a_3}=...=\dfrac{a_{n-1}}{a_n}=\dfrac{a_n}{a_1}\). Và \(a_1+a_2+...+a_n\ne0;a_1=-\sqrt{5}\). Tính \(a_2;a_3;...a_n=?\)
Tính tổng: A= \(\frac{c}{a_1.a_2}+\frac{c}{a_2.a_3}+...+\frac{c}{a_{n-1}.a_n}vớia_2-a_1=a_3-a_2=...=a_n-a_{n-1}=k\)
Mọi người giúp hộ mk bài này nha!!!!!
Cho \(\frac{a_1}{a_2}=\frac{a_2}{a_3}=\frac{a_3}{a_4}=...=\frac{a_{2015}}{a_{2016}}\)
C/minh: \(\left(\frac{a_1+a_2+a_3+...+a_{2015}}{a_2+a_3+a_4+...+a_{2016}}\right)^{2015}=\frac{a_1}{a_{2016}}\)