You finally get an allowance . You put $2 away in January, 4$ away in February, $8 away in March, $16 away in April and followed this saving pattern through to December. How much money do you have in 12 months? Please explain
6 answers
2+4+8+...+2^n = 2^(n+1)-2
2,4,8,16,......
This is a geometric sequence. That means the same number is multiplied, which is 2.
You could calculate this way;
{a, ar, ar^2, ar^3, ar^4, …..}
2, 2x2 2x2^2 2x2^3 2x2^4
(2, 4, 8, 16, 32, ...)
a = the 1st term.
common ratio = 2 > the factor between terms.
Formula:
an = ar^(n - 1)
an = 2 x 2^(n – 1)
a = 2 > 1st term
r = 2 > common ratio
n = nth term > the term you're going to.
The third and fourth term you already know.
a(3)= 2 x 2^(3 - 1)
a(3)= 2 x 2^2
a(3)= 2 x 4
a(3) = $8
a(4)= 2 x 2^(4 - 1)
a(4)= 2 x 2^3
a(4)= 2 x 8
a(4)= $16
a(12)= 2 x 2^(12 - 1)
a(12)= 2 x 2^11
a(12)= 2 x 2048
a(12)= %4096
Answer:
$4096
This is a geometric sequence. That means the same number is multiplied, which is 2.
You could calculate this way;
{a, ar, ar^2, ar^3, ar^4, …..}
2, 2x2 2x2^2 2x2^3 2x2^4
(2, 4, 8, 16, 32, ...)
a = the 1st term.
common ratio = 2 > the factor between terms.
Formula:
an = ar^(n - 1)
an = 2 x 2^(n – 1)
a = 2 > 1st term
r = 2 > common ratio
n = nth term > the term you're going to.
The third and fourth term you already know.
a(3)= 2 x 2^(3 - 1)
a(3)= 2 x 2^2
a(3)= 2 x 4
a(3) = $8
a(4)= 2 x 2^(4 - 1)
a(4)= 2 x 2^3
a(4)= 2 x 8
a(4)= $16
a(12)= 2 x 2^(12 - 1)
a(12)= 2 x 2^11
a(12)= 2 x 2048
a(12)= %4096
Answer:
$4096
Chelle's solution gives the 12th term, but does not show the sum of all 12 terms, which was what was asked.
The answer is 2+4+8+16 = 60. That’s the answer
i got 8190 thank you
yeah 8190 dumbos
3 answers