Answer:
A.
Step-by-step explanation:
[tex]\text{If}\ A(n)\ \text{represents a geometric sequence, then}\ \dfrac{A(n)}{A(n-1)}=\bold{constant}.\\\\A.\\\\A(n)=P(1+i)^{n-1}\\\\A(n-1)=P(1+i)^{n-1-1}=P(1+i)^{n-2}\\\\\dfrac{A(n)}{A(n-1)}=\dfrac{P(1+i)^{n-1}}{P(1+i)^{n-2}}\\\\=(1+i)^{(n-1)-(n-2)}=(1+i)^{n-1-n+2}=(1+i)^1=1+i=\bold{constant}[/tex]
[tex]B.\\\\A(n)=(n-1)(P+i)^n\\\\A(n-1)=(n-1-1)(P+i)^{n-1}=(n-2)(P+i)^{n-1}\\\\\dfrac{A(n)}{A(n-1)}=\dfrac{(n-1)(P+i)^n}{(n-2)(P+i)^{n-1}}=\left(\dfrac{n-1}{n-2}\right)(P+i)^{n-(n-1)}\\\\=\left(\dfrac{n-1}{n-2}\right)(P+i)^{n-n+1}=\left(\dfrac{n-1}{n-2}\right)(P+i)^1\neq\bold{constant}[/tex]
others the same
