我想問有關mathematics 的general term(10分)

We can observe the difference between T(1) and T(2),T(2) and T(3),etc.
99-98=T(1)-T(2)=1
98-96=T(2)-T(3)=2
96-92=T(3)-T(4)=4
We observe that, T(n)-T(n+1)=2^(n-1)
99-2^(1-1)=98
99-2^(1-1)-2^(2-1)=96
99-2^(1-1)-2^(2-1)-2^(3-1)=92
Therefore T(n)=99-2^(n-1)-2^(n-2)-2^(n-3)-......-2^0
T(n)=99-2^(n-1)-2^(n-2)-2^(n-3)-......-2^0
=99-[2^(n-1)+2^(n-2)+2^(n-3)+......+2^0]
=99-[(1-2^n)/(1-2)]
=99-(2^n-1)
=100-2^n

2006-11-02 21:05:04 補充：
Sorry, I did it wrongly in some part.T(n)=99-2^(n-1)-2^(n-2)-2^(n-3)-......-2^0=99-[2^(n-1)十2^(n-2)十2^(n-3)十......十2^0]=99-[1-2^(n-1)]/(1-2)=99-[2^(n-1)-1]=100-2^(n-1)

2006-11-02 21:06:49 補充：
The general term should be 100-2^(n-1)Check:T(1)=100-2^(1-1)=100-1=99T(2)=100-2^(2-1)=100-2=98T(3)=100-2^(3-1)=100-4=96T(4)=100-2^(4-1)=100-8=92

100-n(n+1)/2
=[200-n(n+1)]/2