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=
![{k}^{2} + k](https://tex.z-dn.net/?f=%7Bk%7D%5E%7B2%7D%20%20%2B%20k)
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:
19.9%
Step-by-step explanation:
39.8/200 = 0.199 (move decimal two times)
19.9%
or you can just see what number get 200 to 100, which is by dividing by 2. What you do to the bottom you do to the top so 39.8/200 (divide both sides by 2) = 19.9/100
which is 19.9%.
<u><em>Ace Carlos</em></u>
Answer:
82+25=107 yan po ang answer
If you multiply 2x2 you get 4 and multiply another 2 which is 8. Two answers are 8 and 2x2x2.