Proof involving laws of indices

[James10492]

This came up from a problem involving a fairly simple geometric sequence

40, -20, 10, -5...

the first term in the sequence is 40

the common ratio is -.5 (-1/2)

so the nth term of the sequence can be given by 40 x -.5^n-1

now, it turns out you can write this as (-1)^n-1 x 5/2^n-4


40 x .5^n-1 = 5 x 8 x -.5^n-1

=5 x 2³ x -.5^n-1


.5 = 2^-1 , so


5 x 2³ x (-2^-1)^(n-1)

= 5 x 2³ x -2^1-n

= -1 x 5 x 2³ x 2^(1-n)


now 2³ x 2^1-n = 2^4-n, which gives you your 1/2^(n-4)

but now I am stuck - I don't see how -5 x 1/2^(n-4) = (-1)^n-1 x 5/2^n-4

 

[AvgStudent]

Am I missing something here?
[math]40 \times -.5^{n-1}\neq(-1)^{n-1} \times \frac{5}{2^{n-4}}[/math]

 

[AvgStudent]

[math]\left(-\frac{1}{2}\right)^{n-1}\times 40=\left(-2^{-1}\right)^{n-1}\times2^3\times 5\\ =\left(-2^{1-n}\right)\times2^3\times 5\\ =(-1)^{1-n}\times (2)^{1-n}\times2^3\times 5\\ =(-1)^{1-n}\times2^{1+3-n}\times 5\\ =(-1)^{1-n}\times2^{4-n}\times 5[/math]

 

[pka]

This came up from a problem involving a fairly simple geometric sequence
40, -20, 10, -5... the first term in the sequence is 40
the common ratio is -.5 (-1/2)
so the nth term of the sequence can be given by 40 x -.5^n-1

So the nth term of the sequence is [imath]40\cdot\left(\dfrac{-1}{2}\right)^n[/imath] where [imath]n=0,1,2.\cdots[/imath]

 

[James10492]

Thank you for your replies, it was because I did not realise you could factorise (-2^(1-n)) in that way.

 

[Steven G]

40 x .5^n-1 = 5 x 8 x -.5^n-1 is not true. The left side is always positive while the sign of the right side is oscillating.

Your biggest mistake is writing -.5^n-1. It should be (-.5)^n-1
Do you see that -.5^n-1 is ALWAYS negative for all values of n?
(-.5)^n-1 is negative for all even values of n and positive for odd values of n