Geometric Sub-Sequence

Today I have a question inolving sub-sequences and recursively defined squences. I really like these questions, since when you solve it everything just fits in perfectly. Watch it here:

A sequence a_n is defined as:
a₁ = 11
a_{n + 1} = -0.5a_n + 4.5

b_n is defined for natural n's as b_n = a_n - 3

a) Prove that b_n is a geometric sequence
b) Find the sum of all even positioned terms of b_n.

Let's start with a, of course.

A sequence is a geometric sequence if the ratio of every two consecutive terms is constant. So, for our needs, we can show that b_{n + 1}/b_n is a constant, which proves that b_n is a geometric sequence.

By definition:
b_n = a_n - 3
b_{n + 1} = a_{n + 1} - 3

Using the definition for a_n and a_{n + 1}:
b_n = a_n - 3
b_{n + 1} = -0.5a_n + 4.5 - 3 = -0.5a_n + 1.5

Divide both terms:
(-0.5a_n + 1.5)/(a_n - 3) = -0.5(a_n - 3)/(a_n - 3) = -0.5

The quotient of two consecutive terms is a constant, so b_n is a geometric sequence.

b) Since the ratio of consecutive terms in b_n is -0.5, the ratio of consecutive even-positioned terms is that ratio squared, meaning (-0.5)² = 0.25. Now that we have that quotient, we can find the sum of those even-positioned terms.

The formula for the sum of an infinite converging geometric series is a₁/(1 - q). We have q, and now we need a₁. In this case, the first term (which is given the symbol a₁) is b₂, since it's the first even-positioned term. Let's find it:

b₂ = a₂ + 4.5 - 3

Using the recursive definition:
a_{1 + 1} = -0.5a₁ + 4.5
a₂ = -0.5 * 11 + 4.5 - 3 = -1

So:
b₂ = -1 - 3 = -4

Plug it in the formula:
-4/(1 - 0.25) = -4/0.75 = -4/(3/4) = -16/3

The sum of all even-positioned terms of b_n is -16/3.

If you need math help in anything, go to Super Math Tips, sign up, and you can send any questions you want. They might be answered on this blog!

Hope you like it,
Nadav

nadavs