Respuesta :

chi3fcomm5nd
so even postive integers are by defention in form 2k where k is a natural number so
let the sum of even integers to n=S
S=2(1+2+3+4+5+6+7+8+......+k-1+k
divide bith sides of equation 1 by 2
0.5S=1+2+3+4+5+...........+k-1+k
S=2(k+(k-1)+..............................+2+1)
divide both sides of equation 2 by 2
0.5S=k+k-1+..............................+2+1)
by adding both we will get
___________________________
S=(k+1)(k)
so the sum will be equal to
S=
[tex]{k}^{2} + k[/tex]
so let us test the equation
for the first 3 even number there sums will be
2+4+6=12
by our equation 3^2+3=12
gave us the same answer so our equation is correct

(a) The formula for the sum of the first n even positive integers is n(n+1) and (b) the formula has been proved with an example.

What is an arithmetic progression?

Arithmetic progression is the sequence of numbers that has a fixed common difference between any two consecutive numbers.

For the given situation,

Part (a):

The sum of even numbers formula is obtained by using the sum of terms in an arithmetic progression formula.

Sum of Even Numbers Formula = n(n+1),

where n is the number of terms in the series.

Part (b):

Let us consider the even positive numbers,

2,4,6,8,10,.......

Now take first 5 positive numbers 2,4,6,8,10

By using the formula, n=5

Sum of n even positive integers = [tex]n(n+1)[/tex]

⇒ [tex]5(5+1)[/tex]

⇒ [tex]5(6)[/tex]

⇒ [tex]30[/tex].

Now add the first 5 positive numbers without the formula,

⇒ [tex]2+4+6+8+10=30[/tex]

Hence we can conclude, that the formula for the sum of the first n even positive integers is n(n+1) and the formula has been proved with an example.

Learn more about arithmetic progression here

https://brainly.com/question/19337596

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