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{ }(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`