![61(61+1)=3782](/tpl/images/0378/5075/95b34.png)
The formula for the sum of the first n positive evens is:
. Your question is equivalent to what is the sum of the first 61 even positive integers.
Step-by-step explanation:
122 is the highest even less than 123.
If we solve 2n=122 we can find what number term 122 is in the sequence of positive even numbers.
2n=122
Divide both sides by 2:
n=61
So 122 is the 61st positive even integer. This means we are adding 61 positive even integers.
2(1)+2(2)+2(3)+2(4)+2(5)+2(6)+..........+2(61)
Factor out out 2:
2(1+2+3+4+5+6+...+61)
We could write in summation notation if you prefer:
![2\sum_{k=1}^{61}k](/tpl/images/0378/5075/ffeb3.png)
There is a formula for computing the following:
![\sum_{k=1}^{n}k=\frac{n(n+1)}{2}](/tpl/images/0378/5075/f1e6b.png)
So we have the following:
![2\frac{61(61+1)}{2}](/tpl/images/0378/5075/267f5.png)
![61(61+1)](/tpl/images/0378/5075/4754d.png)
![61(62)](/tpl/images/0378/5075/5ded9.png)
![3782](/tpl/images/0378/5075/82ef2.png)
So if you wanted to know the sum of the first n even numbers it is:
.
Examples:
The sum of the first 4 positive even numbers:
![2+4+6+8=20](/tpl/images/0378/5075/c36df.png)
Now let's put our formula to the test:
![4(4+1)](/tpl/images/0378/5075/06237.png)
![4(5)](/tpl/images/0378/5075/00ff5.png)
![20](/tpl/images/0378/5075/1877e.png)
The sum of the first 10 positive even numbers:
![2+4+6+8+10+12+14+16+18+20=110](/tpl/images/0378/5075/79127.png)
Now let's put out formula to the test again:
![10(10+1)](/tpl/images/0378/5075/34f65.png)
![10(11)](/tpl/images/0378/5075/06578.png)
![110](/tpl/images/0378/5075/939f4.png)
So as you can see the formula works.
Let me know if you want me to actually prove with mathematical induction.