12, 48, 192?

Question

a) common ratio?
b)expression nth term
c) show sum first 30 = 2 to the power of 26-4

oops 62!
thanks for pointing that out

Answer 1

Each term is 4 x the previous term.
F(n) = 4F(n-1) and the common ratio is 4

Next few terms are 768, 3072, 12,288

F(1) = 12 = 16 - 4 = (2^4) - 4

F(1) + F(2) = 60 = 64 - 4 = (2^6) - 4

F(1) + F(2) + F(3) = 252 = 256 - 4 = (2^8) - 4

F(1) + F(2) + F(3) + F(4) = 1020 = 1024 - 4 = (2^10) - 4

F(1) + F(2) + F(3) + F(4) + F(5) = 4092 = 4096 - 4 = (2^12) - 4

F(1) + F(2) + F(3) + F(4) + F(5) + F(6) = 16,380 = 16,384 - 4 = (2^14) - 4

So the sum of the first n terms = [2^(2n +2)] - 4
and the sum of the first 30 terms = [2^(2*30 +2)] - 4 = [2^(62)] - 4

I take it you inadvertently made a typo, transposing the 2 and the 6 when typing them in. It is easily done.

Answer 2

If you look closely you can see that this is a geometric pattern. So with any other geometric pattern you divide the terms to find the common ratio:

48 / 12 = 4.

For the nth term we need to create an expression to represent the pattern. And since it's a geometric, we can use the generic one:

And we get: 3(4)^n

And to get the sum we use the nifty formula:

(first term)*(1-r^n)/(1-r) =
12*(1-4^30)/(1-4) =
-4 *(1-4^30) =
-4 + 4^31 =
-4 + 2^62 =
2^62 - 4

Answer 3

a)
a = 12
ar = 48
r = 48/12

Check:
ar^2 = 192
r = (ar^2)/(ar) = 192/48 = 4
OK
The common ratio is 4

b)
T[1] = a = 12
T[n] = ar^(n-1) = 12*4^(n-1)

c)
S(n) = a(r^n - 1) / (r - 1)
S(2) = 12*(4^2 - 1) / (4 - 1)
= 12*15/3
= 60
12 + 48 = 60

S(30) = 12*(4^30 - 1) / (4 - 1)
= (12 * 2^60 - 12) / 3
= 4 * 2^60 - 4
= 2^62 - 4
= 4,611,686,018,427,387,900

I think you have a typo in the question. The sum is 2^62 - 4 and not 2^26 - 4.

Answer 4

a) 48 / 12 = 4. The ratio between terms is 4.

b) a(n+1) = 4 a(n)
a(n) = 3 * 4^n

c) This is the sum of a finite geometric series.

Sum (low 1 up 30) 3 * 4^n
reindex to start at 0
= Sum (low 0 up 29) 12 * 4^n
= 12 Sum (low 0 up 29) 4^n

Apply the formula for a geometric series.

Sum (low 0 up N) a * r^n = a (1 - r^(N+1)) / (1 - r)

= 12 (1 - 4^30) / (1 - 4)

Negate the terms in parentheses.

= 12 ( 4^30 -1) / 3

Simplify

= 4 ( 4^30 -1)
= 4^31 - 4
= (2^2)^31 - 4

Compound power rule.

= 2^(2 * 31) - 4
= 2^62 - 4

Answer 5

You get the common ratio by dividing any term by the term just before it. So 48/12 = 4

nth term of a geometric sequence = a1(r^(n-1)) where r = common ratio and a1 = first term, so a(n) = 12(4^(n-1))

Sum = a1 times (1 - r^n) over (1 - r)

= 12 (1 - 4^30)/(1 - 4)

= -4(1 - 4^30)

= -4 + 4^31 , and 4 = 2^2 so 4^31 = (2^2)^31 = 2^62

I assume you typed the power in backwards

Answer 6

Common ratio = 48/12= 4

nth term is 3*2^2n

S = 12(1-4^30)/(1-4) = (4^30 -1)4 = 4^31-4 = 2^62 -4

Looks like you reversed 62 in your answer.

Answer 7

Continue summing the digits of each number until you have one digit and it will be 3.

b

Answer 8

a) 4
b) 3*4^n

Answer 9

multiply each number by four