樂高大師介紹 - The Brick Consultant - Steve Guinness
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邱老師在IG上追蹤thebrickconsultant,常常看到很多有趣的樂高創作小品,
今天看到thebrickconsultant發表貼圖,說自己接受Automata Magazine的訪問,
令我好奇地一探究竟,想知道Automata Magazine是甚麼樣的雜誌。
上了Automata M...
1 週前
1 ?
回覆刪除First, we have to find the relation between an and n.
Second, find the limit.
Result : an = (n+1) ^2
an / n^2 = 1 + 2*n^-1 + n^-2
as n tends to infinity, n^-1 and n^-2 tend to 0.
Thus, the final result should be 1.
Hi 小文,
刪除Regarding the an, it isn't equal to (n+1)^2.
It should be
an = (n+1)(2n+1)/2 - n -(2n+1) = (2n^2 - 3n -1)/2
The final result is equal to 1.
First, an must be positive integer.
刪除by putting n=1,n=2 ,n=3 into an = (n+1)(2n+1)/2 - n -(2n+1) = (2n^2 - 3n -1)/2,
we may get a negative number or fraction.
You are right.
刪除a1=0=1*0
a2=2=2*1
a3=6=3*2
a4=12=4*3
:
So actually, an = n(n-1)
You are right.
刪除a1=0=1*0
a2=2=2*1
a3=6=3*2
a4=12=4*3
:
So actually, an = n(n-1)