How Do You Calculate the Sum of a Geometric Sequence with Alternating Terms?

[bold]

I'm given the sequence t(n) = 3 (-1)^n (0.5)^n ; n >= 1

It first asks for the sum of the terms t(1) + t(2) + ... + t(99) which is fine, but it follows up by asking the sum of t(1) + t(3) + t(5) + ... + t(99).

Would i be using partial sums to solve this? If so, i don't know how to find the sum of (-0.5)^2n-1 for n = 1 to 99.

Any help would be appreciated, thanks in advance.

 

[bold]

never mind...

you know, sometimes i wonder about myself...

 

[quentinchin]

Give you a hint...this is a geometric sequence
where 3(-1)^n(0.5)^n = 3(-.5)^n

 

[murshid_islam]

as quentinchin said, this is a geometric sequence
t(n) = 3(-1)^n(0.5)^n = 3(-0.5)^n

now,
t(1) + t(3) + t(5) + ... + t(99)
= 3 [(-0.5)^1 + (-0.5)^3 + (-0.5)^5 + ... + (-0.5)^99 ]

now the series inside the brackets [] has first term, a = (-0.5)^1 = -0.5 and common ratio r = (-0.5)^2 = 0.25

therefore,
3 [(-0.5)^1 + (-0.5)^3 + (-0.5)^5 + ... + (-0.5)^99 ]
= 3 [a/(1-r)]
= 3 [-0.5/(1-0.25)]
= 3 [-0.5/0.75]
= -2

 

[murshid_islam]

i made a little mistake in my last reply.

t(1) + t(3) + t(5) + ... + t(99)
= 3 [(-0.5)^1 + (-0.5)^3 + (-0.5)^5 + ... + (-0.5)^99 ]

now the series inside the brackets [] has first term, a = (-0.5)^1 = -0.5, common ratio r = (-0.5)^2 = 0.25 and number of terms, n = 50

therefore,
3 [(-0.5)^1 + (-0.5)^3 + (-0.5)^5 + ... + (-0.5)^99 ]
= 3 [a(1 - r^n)/(1-r)]
= 3 [-0.5(1-(-0.5)^50)/(1-0.25)]

but (-0.5)^50 is so close to zero that we can ignore that. therefore,

3 [-0.5{1-(-0.5)^50}/(1-0.25)]
= 3[-0.5(1-0)/(1-0.25)]
= 3 [-0.5/0.75]
= -2