Answer to Question #92468 in Algebra for Karla carreras

Question

What is the formula for a geometric series and how can you use it in a practical situation ? Show explanation !

Answer 1

Answer:

A sequence is an ordered list of numbers and series is the sum of a list of numbers (https://www.tutapoint.com/knowledge-center/view/difference-between-sequence-and-series).

The elements of a geometric sequence are a, ar, ...,arⁿ, ..., where "a" represents the initial term and r is the ratio of any two successive elements of the series.

The sum of the finite number of terms is then a+ar+ar²+..+arⁿ = a(1+r+...rⁿ)= a(1-rⁿ⁺¹)/(1-r)

We can prove it using induction by n.

Induction base case n=1: a+ar= a(1+r)= a(1-r²)/(1-r) =a(1-r¹⁺¹)/(1-r) holds

Induction general step

We assume the summation holds for a positive n>1 and we need to prove it also holds for n+1.

Namely, we need to prove that: a+ar+ar²+..+arⁿ⁺¹ = a(1-rⁿ⁺²)/(1-r)

Butd a+ar+ar²+..+arⁿ⁺¹ = (a+ar+ar²+..+arⁿ) +arⁿ⁺¹ = a(1-rⁿ⁺¹)/(1-r) + arⁿ⁺¹ (using the induction hypothesis)

= a[(1-rⁿ⁺¹)/(1-r) + rⁿ⁺¹] =

=a(1-rⁿ⁺¹ + rⁿ⁺¹ - rⁿ⁺²)/ (1-r) = a(1-rⁿ⁺²)/(1-r) holds

Thus, the induction proof is now complete.

Going back to our problem, we have obtained that the formula for a geometric series is a(1-rⁿ⁺¹)/(1-r), when we sum the fist n terms as described above.

There are multiple situations when this formula is useful. For instance, when |r| < 1, rⁿ⁺¹ tends to 0 when n tends to infinity (or when it has a very big value), the sum of the geometric series becomes a/(1-r) only because rⁿ⁺¹ can be neglected.

The same conclusion holds if we sum up to (n-1)th term because in that case rⁿ tends to 0 when n tends to infinity.

For instance, in a production environment, if we know that each day the no. of items is a fraction r of the previous day, then we can estimate the no. of items produced in n days or so.