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=

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
Answer:
Yes, I am 99% sure your right.
Here, DH = HF
x+3 = 3y
x = 3y-3 ----(I)
GH = HE
4x-5 = 2y+3
4x = 2y+8
Substitute value of x,
4(3y-3) = 2y+8
12y-12 = 2y+8
12y-2y = 12+8
10y = 20
y = 2
Now, substitute it in equation 1,
x = 3(2)-3
x = 6-3 = 3
So, your final answer is x=3 & y=2
Answer:
kainis nman pareho tau walang sumagot sa tanong nating dalwa