# Binary System, Amount of Memory (Applications of Geometric Sequences and Series)

Binary System

In a computer, numbers are stored in memory locations where spots are magnetized or not magnetized. These spots can be interpreted as zero or one to represent numbers in a base-two number system. A base-two number system uses the same principle of positional notation as a base-ten number system or the decimal system. A single one or zero in a base-two number system is called a bit which is short for binary digit. Designating a computer as an 8-bit or 16-bit computer indicates that the number of bits that can be stored in a single memory location. The value of each bit read from right to left across the number is given by the corresponding term of a geometric progression with a common ratio of two.

📌 Question 1. What is the value of the fourth bit from the right in the binary number 101111?
✍ Solution:
This is a geometric progression with a1=1 and r=2. So that the value of a4 is

an=a1rn-1
a4=1⋅24-1=23=8.

📌 Question 2. What is the largest number that can be stored in an eight-bit binary number?
✍ Solution:
This is the same as asking for the sum of the geometric progression of 8 terms with a1=1 and r=2. 255 is the largest base-ten number that can be stored in eight bits.

💎 Amount of Memory
📌 Question 3. In 1970 the cost of 1 megabyte of computer memory was \$2025. In 1980 the cost for the same amount of memory had reduced to \$45, and by 1990 the cost had dropped to \$1.
(a) Assuming the pattern continues through the years, what was the cost of 1 megabyte of memory in the year 2000?
(b) How much memory, in megabytes, could you buy for \$10 in the year 2010 based on the trend?
✍ Solution:
(a) (1) Present the given information in a table
(2) Study the table. The information suggests a geometric sequence for the cost at each ten-year interval. Verify this by checking for a constant ratio between successive terms.
45÷2025=1/45 and 1÷45=1/45, so the three terms form a geometric sequence common ratio r=1/45.
(3) To find the cost in the year 2000, find the fourth term in the sequence by multiplying the preceding (third) term by the common ratio. (4) Interpret the result and clearly answer the question.
In the year 2000 you would have paid about 2 cents for a megabyte of memory.

(b) (1) If the cost of 1 megabyte can be found in the year 2010, then the amount of memory purchased for \$10 can be determined. To find the predict cost in the year 2010, the fifth term in the sequence needs to be determined. of a dollar per megabyte.

(2) Take the reciprocal of t5 to get the amount of per dollar.
The amount of memory per dollar is 2025 megabytes.
(3) Find the amount of memory that can be purchased for \$10
So \$10 would buy 10×2025=20,250 megabytes.

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