Difference between revisions of "2015 AMC 8 Problems/Problem 9"
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===Solution 2=== | ===Solution 2=== | ||
| − | The sum is just the first <math>20</math> odd counting/natural numbers, which is <math>20^2=\boxed{\textbf{(D)}~400}</math> | + | The sum is just the first <math>20</math> odd counting/natural numbers, which is <math>20^2=\boxed{\textbf{(D)}~400}</math>. |
Note: The sum of the first <math>x</math> odd numbers is x^2 | Note: The sum of the first <math>x</math> odd numbers is x^2 | ||
Revision as of 07:00, 21 March 2023
Contents
Problem
On her first day of work, Janabel sold one widget. On day two, she sold three widgets. On day three, she sold five widgets, and on each succeeding day, she sold two more widgets than she had sold on the previous day. How many widgets in total had Janabel sold after working 
days?
Solutions
Solution 1
First, we have to find how many widgets she makes on Day . We can write the linear equation 
to represent this situation. Then, we can plug in for 
: 
-- 
-- 
. The sum of 
is 
.
Solution 2
The sum is just the first odd counting/natural numbers, which is 
.
Note: The sum of the first odd numbers is x^2
Solution 3
We can easily find out she makes 
widgets on Day . Then, we make the sum of 
by adding in this way: 
, which include 
pairs of 
. So the sum of 
is 
----LarryFlora
Video Solution
~savannahsolver
See Also
| 2015 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 8 |
Followed by Problem 10 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
