Question

I need help finding the 0th (term before 4) and 15th term.

Answer 1

Series = (-1)^(n+1) * (2)^(4 - 2n)

$Series\colon A_n=\text{ (-1)}^{(n+1)}\cdot2^{(4-2n)}$$A_0=(-1)^{(0+1)}\cdot2^{(4-2\cdot0)}=(-1)\cdot2^4=-16$$A_{15}=(-1)^{(15+1)}\cdot2^{(4-2\cdot15)}=(1)\cdot2^{-26}=\text{ }\frac{1}{67108864}$$A_1=(-1)^{(1+1)}\cdot2^{(4-2\cdot1)}=(-1)^2\cdot2^2=\text{ 1}\cdot4=4$$A_2=(-1)^{(2+1)}\cdot2^{(4-2\cdot2)}=(-1)^3\cdot2^0=(-1)(1)\text{ = -1}$$A_3=(-1)^{(3+1)}\cdot2^{(4-2\cdot3)}=(-1)^4\cdot2^{(4-6)}=(1)\cdot2^2=4$