我有2個數學問題.請大家幫幫忙

1.
A2 = A1
A3 = A1+2*A2
A4 = A1+2*A2+3*A3
A5 = A1+2*A2+3*A3+4*A4


2.
A(n+1) = (An+3)/2
A1 = 1, 故
A2 = (1+3)/2 = 2
A3 = (2+3)/2 = 5/2
一般:
An = (A(n-1)+3)/2
= A(n-1)/2 + 3/2
= (A(n-2)+3)/2^2 + 3/2
= A(n-2)/2^2 + 3/2^2 + 3/2
= A(1)/2^{n-1} + 3/2^{n-1}+...+3/2^2+3/2
= 1/2^{n-1} + 3/2^{n-1}+...+3/2^2+3/2
= 1/2^{n-1} + (3/2)[1-(1/2)^{n-1}]/(1-1/2)
= 3 - 3/2^{n-1}+1/2^{n-1}
= 3 - 2/2^{n-1} = 3-1/2^{n-2} = 3-4/2^n.