Question

Please help! Will give medal!
Find the first six terms of the sequence.

a1 = 1, an = 4 • an-1

Answer 1

a1 = 1
a2 = 4* a(2-1)
a2 = 4*a1 = 4*1 = 4
a3 = 4* a(3-1)
a3 = 4*a2 = 4*4 = 16
a4 = 4* a(4-1)
a4 = 4*a3 = 4*16 = 64
and so on for 5th and 6th term

Answer 2

thank you so much! @sangya21

Answer 3

anytime :)

Answer 4

would you be able to help me with another one if possible?

Answer 5

okkk

Answer 6

Write the sum using summation notation, assuming the suggested pattern continues.

-8 - 3 + 2 + 7 + ... + 67

Answer 7

a = -8
d = 5
an = 67
67 = a +(n-1)d
67 = -8 +(n-1)5
75 = (n-1)*5
15 = n-1
n = 16
\[\sum_{k=0}^{n} \space= \space n/2\space (2a +(n-1)d)\]
put n = 16
d = 5
a = -8

Answer 8

i plugged it in but it didnt make sense

Answer 9

n/2(2a +(n-1)d)
16/2 (2*-8 +(16-1)*5)
8(-16 +(15*5))
8(-16 +75)
8*59
472