Give a recursive and non-recursive equation for the following sequence:{3,-12,48,-192,...}

Question

Give a recursive and non-recursive equation for the following sequence:{3,-12,48,-192,...}

Answer 1

This is a geometric sequence. The formula is l=a(r)^(n-1) where l is the last term, a is the first term,

r is the common ratio, and n is the number of terms.

{3, -12, 48, -192, ...}

To find any term use 3[(-4)^(n-1)]

The first term is 3[(-4)^(1-1)]=3(-4)^0=3(1)=3

The second term is 3[(-4)^(2-1)]=3[(-4)^1]=3(-4)=-12

The third term is 3[(-4)^(3-1)]=3[(-4)^2]=3(16)=48

The fourth term is 3[(-4)^(4-1)]=3[(-4)^3]=3[-64)=-192

The fifth term is 3[(-4)^(5-1)]=3[(-4)^4]=3(256)=768

Looking at the sequence, multiply any term by -4 to

get the next term in the sequence.

 

Answer 2

Hi Louise;

3(-4)=-12

-12(-4)=48

48(-4)=-192

RECURSIVE

a₁=3

a_n=(-4)a_n-1

NON-RECURSIVE

y=-4x