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Sequence and Series

Geometric Series

Nabil Akbarazzima Fatih

Mathematics

Concept of Geometric Series

Consider the data on the number of Covid-19 infected patients in the following table:

MonthJanuaryFebruaryMarchAprilMay
Number of patients41236108324

The data above shows a pattern of increasing numbers of patients each month. If we sum the number of patients from the first month up to a certain month, we form a Geometric Series.

A Geometric Series is the sum of the terms of a geometric sequence. A geometric sequence is a sequence of numbers where the ratio between any two consecutive terms is always constant. This ratio is called the common ratio denoted by (r)(r).

In the Covid-19 patient data:

  • The first term (aa or U1U_1) is 4.
  • The ratio (r)(r) = 124=3612=10836=324108=3\frac{12}{4} = \frac{36}{12} = \frac{108}{36} = \frac{324}{108} = 3.

So, the sequence of patient numbers is 4,12,36,108,3244, 12, 36, 108, 324. The geometric series is the sum of the terms of this sequence:

  • Sum of the first 2 months (S2)(S_2) = 4+12=164 + 12 = 16
  • Sum of the first 3 months (S3)(S_3) = 4+12+36=524 + 12 + 36 = 52
  • Sum of the first 4 months (S4)(S_4) = 4+12+36+108=1604 + 12 + 36 + 108 = 160
  • and so on.

Finding the Formula for the Sum of the First n Terms

How can we calculate the sum of the first nn terms (Sn)(S_n) without adding them one by one? Let's find the formula.

Consider this table which shows the process of rediscovering the formula for the sum of a geometric series:

NotationDirect SummationUsing Un+1U_{n+1} and U1U_1General Form
S2S_24+12=164 + 12 = 16S2=36−43−1=322=16S_2 = \frac{36 - 4}{3 - 1} = \frac{32}{2} = 16S2=U3−U1r−1S_2 = \frac{U_3 - U_1}{r - 1}
S3S_34+12+36=524 + 12 + 36 = 52S3=108−43−1=1042=52S_3 = \frac{108 - 4}{3 - 1} = \frac{104}{2} = 52S3=U4−U1r−1S_3 = \frac{U_4 - U_1}{r - 1}
S4S_44+12+36+108=1604 + 12 + 36 + 108 = 160S4=324−43−1=3202=160S_4 = \frac{324 - 4}{3 - 1} = \frac{320}{2} = 160S4=U5−U1r−1S_4 = \frac{U_5 - U_1}{r - 1}
⋮\vdots…\dots…\dots…\dots
SnS_n…\dots…\dotsSn=Un+1−U1r−1S_n = \frac{U_{n+1} - U_1}{r - 1}

From the bottom right of the table, we get the general form:

Sn=Un+1−U1r−1S_n = \frac{U_{n+1} - U_1}{r - 1}

We know that the formula for the nn-th term in a geometric sequence is Un=arn−1U_n = ar^{n-1}. Thus, Un+1=ar(n+1)−1=arnU_{n+1} = ar^{(n+1)-1} = ar^n. Substitute Un+1=arnU_{n+1} = ar^n and U1=aU_1 = a into the SnS_n formula:

Sn=arn−ar−1S_n = \frac{ar^n - a}{r - 1}
Sn=a(rn−1)r−1S_n = \frac{a(r^n - 1)}{r - 1}

This is the formula for the sum of the first nn terms of a geometric series.

Geometric Series Formula

In general, the formula to calculate the sum of the first nn terms of a geometric series is:

Sn=a(rn−1)r−1, for r>1S_n = \frac{a(r^n - 1)}{r - 1}, \text{ for } r > 1

or

Sn=a(1−rn)1−r, for r<1S_n = \frac{a(1 - r^n)}{1 - r}, \text{ for } r < 1

Legend:

  • SnS_n = sum of the first nn terms
  • aa = first term (U1)(U_1)
  • rr = ratio (r≠1)(r \neq 1)
  • nn = number of terms

Example Application

A bicycle company's production in 2020 increased monthly, forming a geometric sequence. Production in January was 120 units. In April, production reached 3,240 units. What was the total bicycle production up to May?

Solution:

  • January Production (U1)(U_1) = a=120a = 120
  • April Production (U4)(U_4) = 3,240
  • Question: Total production up to May (S5)(S_5)

Step 1: Find the ratio (r)(r)

U4=ar4−1=ar3U_4 = ar^{4-1} = ar^3
3.240=120⋅r33.240 = 120 \cdot r^3
r3=3.240120r^3 = \frac{3.240}{120}
r3=27r^3 = 27
r=273=3r = \sqrt[3]{27} = 3

Step 2: Calculate S5S_5. Since r=3>1r = 3 > 1, use the formula:

Sn=a(rn−1)r−1S_n = \frac{a(r^n - 1)}{r - 1}
S5=120(35−1)3−1S_5 = \frac{120(3^5 - 1)}{3 - 1}
S5=120(243−1)2S_5 = \frac{120(243 - 1)}{2}
S5=120(242)2S_5 = \frac{120(242)}{2}
S5=60⋅242S_5 = 60 \cdot 242
S5=14.520S_5 = 14.520

Therefore, the total bicycle production up to May is 14,520 units.

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Infinite Geometric Series