Question

Find a formula for the sequence:

Answer 1

\[\left\{ -1, \frac{ 2 }{ 3 }, -\frac{ 1 }{ 3 }, \frac{ 4 }{ 27 },- \frac{ 5 }{ 81 } \right\}\]

Answer 2

I understand the alternating negatives, but the third term is throwing me off as far as any relationship between the numerators

Answer 3

Would it help to find the LCD here?

Answer 4

@jim_thompson5910 @zepdrix @AloneS @pooja195

Answer 5

I rewrote this as:

\[\left\{- \frac{ 81 }{ 81 }, \frac{ 54 }{ 81 }, -\frac{ 27 }{ 81 }, \frac{ 12 }{ 81 }, -\frac{ 5 }{ 81 } \right\}\]

Didnt really help though

Answer 6

\[\Large \left\{ -1, \frac{ 2 }{ 3 }, -\frac{ 1 }{ 3 }, \frac{ 4 }{ 27 },- \frac{ 5 }{ 81 } \right\}\]
is equivalent to
\[\Large \left\{ -\frac{1}{1}, \frac{ 2 }{ 3 }, -\frac{ 3 }{ 9 }, \frac{ 4 }{ 27 },- \frac{ 5 }{ 81 } \right\}\]
I changed the -1 to -1/1
I changed the -1/3 to -3/9

Answer 7

Oh that looks a lot better

Answer 8

the numerators (1,2,3,4,5) are incrementing by 1

the denominators are powers of 3
1 = 3^0
3 = 3^1
9 = 3^2
27 = 3^3
81 = 3^4

also, the terms are alternating in sign, so you'll need a (-1)^n or (-1)^(n+1) term in there

Answer 9

so (n-1)^3 for the denominator?

Answer 10

no, thats not right. The denominator is confusing me, Im not seeing how to relate it to n.

Answer 11

no, not cubed, powers of 3

Answer 12

x^3 is different from 3^x

Answer 13

the denominator would be 3^(n-1) if you start n at n = 1

Answer 14

\[a _{n}=\frac{ n }{ 3^{(n-1)} }(-1^{n})\]

Answer 15

I think this is the correct answer

Answer 16

it's very close, but not quite there

Answer 17

you need to write (-1^n) as (-1)^n

Answer 18

n needs to be outside the parentheses?

Answer 19

sweet ok, gotcha. Thanks for your help Jim

Answer 20

consider when n = 2

(-1^n) = (-1^2) = (-1*1^2) = -1
while
(-1)^n = (-1)^2 = (-1)*(-1) = 1

Answer 21

no problem