Find a formula for the sum of the first n even positive integers.
1 answer:
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
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
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Answer:
k = -6.29
Step-by-step explanation:
91 - 2 + 7k + 1 = 46
add all like terms
90 + 7k =46
subtract
90 + 7k = 46
-90 -90
------------------------
7k = -44
divide
=
(these arent fractions thats just how i divide things :))
= -6.28571428571
round to nearest hundredth
k= -6.29
goes up to 9 because 5 makes 8 go up one when rounding
sorry for asking so many questions. hope this helps
You need to find the change in x and y
change in x = 12
change in y = 16
to find the distance you use a^2+ b^2 = c^2
or in this case x^2 + y^2 = d^2
144+256 = 400
=20
so distance equals 20
I hope this helps!
change in x = 12
change in y = 16
to find the distance you use a^2+ b^2 = c^2
or in this case x^2 + y^2 = d^2
144+256 = 400
=20
so distance equals 20
I hope this helps!