Dear Mr. Yue,
I would like to attempt question no. 1,3,5
1. For n points in the plane. Point a_1 has linkage with remaining n-1 points, same for a_2…,
same for a_2, therefore there’s n(n-1)/2 sides in total. After subtracting the no. of sides (n) from
n(n-1)/2, the answer could be obtained: n(n-1)/2 –n, or n(n-3)/2
3. The no. of ways is the same of the numbers in the Pascal Triangle, so the answer for
A is 5C0=1, B is 5C1=5, C is 5C2=10, D is 5C3=10, E is 5C4=5, F is 5C5=1.
5. Let a_n for the no. of ways the student could jump up the stairway,
When I take a step forward, there 's a_n-1 ways for the student to jump up for the remaining
steps
When I take two steps forward, there's a_n-2 ways for the student to jump up for the remaining
steps, then a recurrence relation is setup,
a_n=a_n-1+a_n-2
a_1 =1 because there's only one way for the student to jump up a stairway with one step,
a_2=2 because there's only two ways for the student to jump up a stairway with two steps,
a_3=1+2=3
a_4=2+3=5
a_5=8, a_6=13, a_7=21, a_8=34, a_9=55, a_10=89
For the general case,
Consider the characteristic equation,
A^n=A^(n-1)+ A^(n-2)
A^2-A-1=0
By the quadratic Formula,
A=
1 5
1 5 n
1 5 n
)  y(
)
, let a_n = x(
2
2
2
When n=1, a_1=1, When n=2, a_2=2. After solving the simultaneous equation, we have:
A_n=
n
n
1  1  5   1  5  

 
 
5  2   2  

