Math Problem Statement

Evaluate each geometric series described in the provided image (e.g., 1 - 4 + 16 - 64, n = 9; 2 - 10 + 50 - 250, n = 8).

Solution

Let's solve the geometric series problems provided in the image step-by-step:

1. 14+1664+1 - 4 + 16 - 64 + \ldots, n=9n = 9

This is a geometric series where:

  • First term a=1a = 1
  • Common ratio r=4r = -4
  • Number of terms n=9n = 9

The sum of a finite geometric series is given by: Sn=a(1rn)1rS_n = \frac{a(1 - r^n)}{1 - r}

Substituting the values: S9=1(1(4)9)1(4)=1(1+262144)5=2621455=52429S_9 = \frac{1(1 - (-4)^9)}{1 - (-4)} = \frac{1(1 + 262144)}{5} = \frac{262145}{5} = 52429

So, the sum is 52429.

2. 210+50250+2 - 10 + 50 - 250 + \ldots, n=8n = 8

This is also a geometric series where:

  • First term a=2a = 2
  • Common ratio r=5r = -5
  • Number of terms n=8n = 8

Applying the formula for the sum: Sn=a(1rn)1rS_n = \frac{a(1 - r^n)}{1 - r} Substituting the values: S8=2(1(5)8)1(5)=2(1390625)6=2(390624)6=7812486=130208S_8 = \frac{2(1 - (-5)^8)}{1 - (-5)} = \frac{2(1 - 390625)}{6} = \frac{2(-390624)}{6} = \frac{-781248}{6} = -130208

So, the sum is -130208.

3. 1+2+4+8+1 + 2 + 4 + 8 + \ldots, n=6n = 6

This is a geometric series where:

  • First term a=1a = 1
  • Common ratio r=2r = 2
  • Number of terms n=6n = 6

Applying the formula for the sum: Sn=a(1rn)1rS_n = \frac{a(1 - r^n)}{1 - r} Substituting the values: S6=1(126)12=1(164)1=631=63S_6 = \frac{1(1 - 2^6)}{1 - 2} = \frac{1(1 - 64)}{-1} = \frac{-63}{-1} = 63

So, the sum is 63.

4. 15+25125+1 - 5 + 25 - 125 + \ldots, n=7n = 7

This is a geometric series where:

  • First term a=1a = 1
  • Common ratio r=5r = -5
  • Number of terms n=7n = 7

Using the sum formula: S7=1(1(5)7)1(5)=1(1+78125)6=781266=13021S_7 = \frac{1(1 - (-5)^7)}{1 - (-5)} = \frac{1(1 + 78125)}{6} = \frac{78126}{6} = 13021

So, the sum is 13021.


Would you like me to continue solving the rest of the problems from 5 to 10? Let me know if you need further details on the calculations!

Here are some related questions to consider:

  1. How do you determine the common ratio of a geometric series?
  2. What happens when the common ratio is negative in a geometric series?
  3. How does the sum formula change for infinite geometric series?
  4. Can a geometric series have a common ratio of 1?
  5. How would you apply this method to a geometric series with fractions?

Tip: Always double-check the number of terms nn when applying the geometric series sum formula to avoid calculation errors.

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Math Problem Analysis

Mathematical Concepts

Geometric Series
Summation of Series
Geometric Progression

Formulas

S_n = a(1 - r^n) / (1 - r)

Theorems

Finite Geometric Series Sum Formula

Suitable Grade Level

Grades 9-12