What are the first four terms of a geometric sequence with t1 = 4 and tn = -3tn-1?

Question

anonymous · Answer

you can follow this example Assuming the recursive definition is tn = 2*tn-1
t1 = 3
t2 = 2*t1 = 2*3 = 6
t3 = 2*t2 = 2*6 = 12
t4 = 2*t3 = 2*12 = 24

anonymous · Answer

ar^0+ar^1..... ar^n
t1        t2             tn

anonymous · Answer

can you draw  

-3tn-1   for us please

anonymous · Answer

What do you mean?

campbell_st · Answer

is it

$$t_{n} = -3t^{n -1}$$

anonymous · Answer

Yeah except the n-1 is not in exponential form...I don't know what it's called haha but I think you read it as -3t(sub)n-1 if that makes sense :p

campbell_st · Answer

ok... so is it 

$$t_{n} = -3t_{n-1}$$

anonymous · Answer

yupp

campbell_st · Answer

ok... that should help @timo86m

campbell_st · Answer

so the pattern is 

4,  4*(-3), (4*-3)*-3, and (4*-3*-3)*-3

campbell_st · Answer

just evaluate for the terms

campbell_st · Answer

so the pattern is saying multiply -3 by the previous turn after starting at 4

ist term is 4

so the 2nd term is 4 times -3  

          3rd term is -3 times the 2nd term

         4th term is -3 times the 3rd term

anonymous · Answer

Thank you! (: