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You finally get an allowance . You put $2 away in January, 4$ away in February, $8 away in March, $16 away in April and followe...Asked by kareema
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
Answers
Answered by
Steve
2+4+8+...+2^n = 2^(n+1)-2
Answered by
Chelle
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
Answered by
Steve
Chelle's solution gives the 12th term, but does not show the sum of all 12 terms, which was what was asked.
Answered by
Lili
The answer is 2+4+8+16 = 60. That’s the answer
Answered by
jeohfkuwahd
i got 8190 thank you
Answered by
bruh
yeah 8190 dumbos
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