## 2015年7月15日 星期三

### 訓練數學感 70 ─ 格子點

#### 5 則留言:

1. 1 ?
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.

1. Hi 小文，

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.

2. First, an must be positive integer.
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.

3. You are right.
a1=0=1*0
a2=2=2*1
a3=6=3*2
a4=12=4*3
:

So actually, an = n(n-1)

4. You are right.
a1=0=1*0
a2=2=2*1
a3=6=3*2
a4=12=4*3
:

So actually, an = n(n-1)