S1 Sequences

a) a[0] = 1, a[n] = (-1)^(n-1) * a[n-1]+1 for all n >= 1

即係個 sequence 係 1,2,3,4 咁去.. 只係逢雙數係負數...

b) a[0] = 4, a[1] = 7, a[n] = a[n-2] + a[n-1] for all n >= 2

即係由第三個數開始，係由之前o個兩個數加埋...
11 = 4 + 7
18 = 7 + 11
29 = 11 + 18
...

a.７
b.１２３

a.f(1)=1=1x(-1)^0
f(2)=-2=2x(-1)^1
f(3)=3=3x(-1)^2
f(4)=-4=4x(-1)^3

so f(n)=n(-1)^n-1

b.this is the Fibonacci Sequence

f(1)=4
f(2)=7
f(3)=f(1)+f(2)=4+7=11
f(4)=f(2)+f(3)=7+11=18

so f(n)=f(n-2)+f(n-1)

Qa.
分2組睇: 1, 3, 5, 7,9.....
-2, -4, -6, -8, -10........
所以 Ans 係: 1, -2, 3, -4, 5, -6, 7, -8, 9, -10, 11, -12.......

Qb.
4+7 =11
7+11 =18
11+ 18 = 29.....
so Ans is: 4, 7, 11, 18, 29, 47, 76, 123, 199......

明唔明???