+ / -檢視主題
由 --- 於 星期三 五月 21, 2003 9:52 pm
反正這種階差題就是相減嘛
由 scsnake 於 星期三 五月 21, 2003 9:43 pm
genius...
由 ---- 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
由 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
由 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？
→這怎麼來的？
由 ---- 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
(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
由 ---- 於 星期三 五月 21, 2003 8:18 pm
i can now
由 --- 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,...
Consider them from n=1,n=2,...
[問題]General Term
由 ---- 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?
(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?