$\left\{\begin{matrix} u_1=-1;u_2=-2 & \\ n.u_{n+2}-(3n+1)u_{n+1}+(2n+2)u_n=3 & \end{matrix}\right.$
#1
Posted 04-09-2015 - 15:48
$\left\{\begin{matrix}
u_1=-1;u_2=-2 & \\ n.u_{n+2}-(3n+1)u_{n+1}+(2n+2)u_n=3 &
\end{matrix}\right.$
với n nguyên dương.
#2
Posted 04-09-2015 - 16:34
ìm số hạng tổng quát của dãy số cho bởi:
$\left\{\begin{matrix}
u_1=-1;u_2=-2 & \\ n.u_{n+2}-(3n+1)u_{n+1}+(2n+2)u_n=3 &
\end{matrix}\right.$
với n nguyên dương.
1 bài thôi bạn ơi http://diendantoanho...ht/#entry587137
"I am the bone of my sword,
Unknown to Death, Nor known to Life,
So as I pray, unlimited blade works."
#3
Posted 04-09-2015 - 16:59
Từ gt ta có: $n(u_{n+2}-2u_{n+1})=(n+1)(u_{n+1}-2u_{n})+3$
<=> $\frac{u_{n+2}-2u_{n+1}}{n+1}=\frac{u_{n+1}-2u_{n}}{n}+\frac{3}{n(n+1)}$
=> $\frac{u_{n+2}-2u_{n+1}}{n+1}=\frac{u_{2}-2u_{1}}{1}+ \frac{3}{1.2}+\frac{3}{2.3}+...+\frac{3}{n(n+1)}$
<=> $\frac{u_{n+2}-2u_{n+1}}{n+1}= 3(1- \frac{1}{n+1})$
<=> $u_{n+2}-2u_{n+1}=3n$
<=> $u_{n+2}+3(n+2)=2[u_{n+1}+3(n+1)]$
=> $u_{n+2}+3(n+2)=2^{n+1}[u_{1}+3]$
$= 2^{n+2}$
Vậy $u_{n}=2^{n}-3n$.
Edited by huypham2811, 04-09-2015 - 17:04.
- dark templar, pcfamily and Belphegor Varia like this
#4
Posted 04-09-2015 - 19:50
Mấy bài kiểu này có thể SD PT sai phân
Tham khảo tài liệu này:
www.MATHVN.com-Day-so-NguyenTatThu.pdf 497.83KB 86 downloads
PP SAI PHÂN DÃY SỐ.pdf 371.43KB 820 downloads
PT SAI PHAN CONG THUC TONG QUAT DAY SO-www.MATHVN.com.pdf 947.53KB 92 downloads
www.MATHVN.com-CTTQ-Dayso-TranDuySon.pdf 157.67KB 919 downloads
Mabel Pines - Gravity Falls
1 user(s) are reading this topic
0 members, 1 guests, 0 anonymous users