Respuesta :

Answer:

[tex]61(61+1)=3782[/tex]

The formula for the sum of the first n positive evens is:

[tex]n(n+1)[/tex]. 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:

[tex]2\sum_{k=1}^{61}k[/tex]

There is a formula for computing the following:

[tex]\sum_{k=1}^{n}k=\frac{n(n+1)}{2}[/tex]

So we have the following:

[tex]2\frac{61(61+1)}{2}[/tex]

[tex]61(61+1)[/tex]

[tex]61(62)[/tex]

[tex]3782[/tex]

So if you wanted to know the sum of the first n even numbers it is:

[tex]n(n+1)[/tex].

Examples:

The sum of the first 4 positive even numbers:

[tex]2+4+6+8=20[/tex]

Now let's put our formula to the test:

[tex]4(4+1)[/tex]

[tex]4(5)[/tex]

[tex]20[/tex]

The sum of the first 10 positive even numbers:

[tex]2+4+6+8+10+12+14+16+18+20=110[/tex]

Now let's put out formula to the test again:

[tex]10(10+1)[/tex]

[tex]10(11)[/tex]

[tex]110[/tex]

So as you can see the formula works.

Let me know if you want me to actually prove with mathematical induction.