Forgot password?
 Register account
View 1265|Reply 1

[数列] 一道数列不等式

[Copy link]

3159

Threads

7941

Posts

610K

Credits

Credits
63770
QQ

Show all posts

hbghlyj Posted 2020-6-12 13:09 |Read mode
数列$\{a_n\}$满足$a_1=1,a_{n}=a_{n-1}+a_{\lfloor \frac n2 \rfloor},$求证$\forall n\in\mathbf N_+,a_{2^n}>2^{\frac{(n+3)(n-1)}4}$

3159

Threads

7941

Posts

610K

Credits

Credits
63770
QQ

Show all posts

 Author| hbghlyj Posted 2020-6-12 13:19
先导一些$a_{2^n}$的表示,也许有用
$a_{2^n}=a_{2^n-1}+a_{2^{n-1}}=a_{2^n-1}+a_{2^{n-1}-1}+a_{2^{n-2}}=\cdots=a_{2^n-1}+a_{2^{n-1}-1}+\cdots+a_{2^2-1}+a_1+a_1$
这个跨步比较大,换成下面的可能会好一些
$a_{2^n}=a_{2^n-2}+a_{2^{n-1}-1}+a_{2^{n+1}}=a_{2^n-3}+2a_{2^{n-1}-1}+a_{2^{n-1}}=a_{2^{n-1}}+2a_{2^{n-1}-1}+2a_{2^{n-1}-2}+\cdots+a_2+2a_1+1+1$

Mobile version|Discuz Math Forum

2025-5-31 10:34 GMT+8

Powered by Discuz!

× Quick Reply To Top Edit