[問題]General Term

[問題]General Term

---- 於 星期三 五月 21, 2003 6:31 pm


If for any natural number n, a_n >0 and
(a_1)^3 + (a_2)^3 +... + (a_n)^3 = (a_1 + a_2 +...+ a_n)^2
Find the general term a_n.

I know that
(a_n)^2 = 2(a_1+a_2+...+a_(n-1)) + a_n
but, what next?

----
訪客
 

--- 於 星期三 五月 21, 2003 8:16 pm


You can do it. You try again.
Consider them from n=1,n=2,...

---
訪客
 

---- 於 星期三 五月 21, 2003 8:18 pm


i can now

----
訪客
 

---- 於 星期三 五月 21, 2003 8:20 pm


(a_n)^2 = 2(a_1+a_2+...+a_(n-1)) + a_n  ...(1)
(a_(n+1))^2 = 2(a_1+a_2+...+a_n) + a_(n+1) ...(2)

(2)-(1)
(a_(n+1))^2 - (a_n)^2 = 2a_n + a_(n+1) - a_n
(a_(n+1))^2 - a_(n+1) = (a_n)^2 + a_n
(a_(n+1))^2 - a_(n+1) + 1/4 = (a_n)^2 + a_n + 1/4
a_(n+1) - 1/2 = a_n + 1/2
a_(n+1) = a_n + 1

When n =1, sub. into the original formula
(a_1)^3 = (a_1)^2
a_1=1
Therefore a_n=n

----
訪客
 

scsnake 於 星期三 五月 21, 2003 9:31 pm


(a_n)^2 = 2(a_1+a_2+...+a_(n-1)) + a_n
→這怎麼來的?

scsnake
訪客
 

Raceleader 於 星期三 五月 21, 2003 9:37 pm


I think ai=j, where i,j=1,2,3,...,n, i can be equal to j

Raceleader
訪客
 

---- 於 星期三 五月 21, 2003 9:42 pm


scsnake 寫到:(a_n)^2 = 2(a_1+a_2+...+a_(n-1)) + a_n
→這怎麼來的?


(a_1)^3 + (a_2)^3 +... + (a_n)^3 = (a_1 + a_2 +...+ a_n)^2...(1)
(a_1)^3 + (a_2)^3 +... + (a_(n-1))^3 = (a_1 + a_2 +...+ a_(n-1))^2...(2)

(2)-(1)得之

----
訪客
 

scsnake 於 星期三 五月 21, 2003 9:43 pm


genius...

scsnake
訪客
 

--- 於 星期三 五月 21, 2003 9:52 pm


反正這種階差題就是相減嘛

---
訪客
 




代數學