Thanks to everyone who entered our Kirk Weiler Voicemail Greeting Contest! We were thrilled with the participation and delighted by so many correct answers to this math problem last month! We originally said there would only be one winner, but after such an enthusiastic turnout during the season of giving, we decided to select FOUR VOICEMAIL CONTEST WINNERS. Congratulations to teacher Carolyn and students Dean, Autumn, and Karishma, who have each received a personalized recording from Kirk to use as their voicemail greeting!
Kirk wanted to offer a math problem that could potentially be solved by everyone from talented middle schoolers to Algebra II students — using either simple but lengthy calculations or some sophisticated math. He landed on the following:
What is the value of the sum: 1+3+5+7+…+99+101?
The answer is 2,601.
Here they are and how they solved it:
This is an arithmetic sequence so I added the first and last terms and multiplied but half of how many terms there are.
Sum of 51 terms: Add the first and last terms (1+101) and multiply by half of how many terms (51/2)
I use eMATH to teach Algebra2. I’m on the site daily and I love the banks of multiple choice questions! The Regents reviews are fantastic!Carolyn, teacher (with excellent taste in T-shirts)
I found the answer by using the Arithmetic Series Formula. Which is Sn=n(a1+a2 over 2).
The sequence was odd, so I put the numbers 1, 3, 5, and 7 in a google sheet and then dragged it down to the number 51 because that was where 101 was. So that determined how I got the n or number terms being added. So n= 51
a1 is the first term in the sequence. So in this problem, it was 1. So a1=1.
a2 is the last term shown in the sequence. So in this problem, it was 101. So a2=101.
Once I found all of the terms, I substituted them into the formula being used which is Sn=n(a1+a2/2).
Now Sn=51(1+101 over 2)
Add 101+1 in the parenthesis which equals 102
Take the 102 and divide it by 2 which equals 51
Now you have Sn=51(51) or 51 squared which equals 2,601 which is the final answer. Sn=2,601
What I love about eMATHinstruction is that no matter what you do in class whether it be Algebra I, Geometry, or Algebra II. You have a site dedicated to making the student understand the subject. I personally struggled in Geometry during the Covid-19 crisis. But at the end of the year, I passed both the class AND the final! If it weren’t for eMathinstruction that year, I would’ve failed due to online learning making me struggle in certain topics such as Geometry where I had a lifeline, or Kirk Weiler and his team at eMathinstruction. Kirk became an inspiration of mine due to his help to many students such as myself. I try to help others myself in subjects beyond math. But I always make sure to throw in a joke about trig just because of Kirk and eMATHinstruction. THANK YOU!Dean, student
First, you need to find the number of terms, so I do 101=1+(n-1)2. This would come out to 51=the number if terms. Then you do S51 = 51/2 (1+101) which would end up being equal to 2,601. So the sum is 2,601.
I love eMATHinstruction because without it, I would know nothing about math. My class still uses eMATHinstruction to learn math. It has been very helpful to help me understand the curriculum and has prepared me for the NYS regents exam in past years. I’m sure it will prepare me again for the algebra 2 regents in June. I also love Kirk’s jokes. They make me laugh.Autumn, student
n = 51
a=1 (starting number)
Arithmetic sequence so Sn = (n/2)(2a+(n-1)d)
Sn = (51/2)(2+(50)(2)) = 2,601
Sum of the sequence is 2,601.
I have watched eMATHinstructions since Algebra 1 and it has truly allowed my mathematics to flourish. More specifically, Kirk’s lesson videos and reviews helped me immensely throughout the years. After a simple review video of Algebra 1, I achieved 100 on the regents. I couldn’t have done it without Kirk or eMATHinstruction.Karishma, student
Stay tuned for our next contest — if you keep thinking and keep solving problems, you could be our next winner!