1 99 100
3 97 100
5 95 100
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Lesson 4-16 for exercise sets.
1 3 5 7 ... 93 95 97 99
7 93 100
Notice that the sums are perfect squares. In fact, each sum
is the square of the number of terms being added.
1 12
1 3 22
1 3 5 32
1 3 5 7 42
1 3 5 7 9 52
Extending this pattern, you find that the sum of all 50 terms will be 50^2 , or 2500.
The total of the perimeters, therefore, is 4(2500 feet) 10,000 feet.
Method 2:Organize Data
As you did when using Method 1, first find the sum of the first 50 odd numbers.
1 3 5 7 9 11 ... 89 91 93 95 97 99
Adding these numbers in order could take a long time. However, if
you look carefully, you can find a way to organizethe terms to make
computing the sum very simple. Notice that the sum of the first term
and the last term, 1 99, is 100. The sum of the second term and the
second to the last term, 3 97, is also 100. You can pair up all the
terms in this way to form sums of 100.
There are 50 odd numbers from 1 to 99, so there are 25 pairs,
each with a sum of 100. So the total is 25(100), or 2500.
The sum of one set of side lengths (that is, the length of one
side of each of the 50 squares) is 2500 feet.
As you did when using Method 1, you can find the total perimeter
by multiplying this sum by 4: 4(2500 feet) 10,000 feet. The total
perimeter is 10,000 feet.
Check to make sure your answer makes sense.
Because you found the same solution using two different strategies,
you can be reasonably certain the solution is correct. If you wish,
you can check by computing 1 3 5 ... 99 on a calculator
and multiplying the result by 4.
