Respuesta :
You can make a start by putting together an expression for the sum of the even integers between 1 and k inclusive.
Let S be the sum of the even integers between 1 and k inclusive.
Then:
S=2+4+6+⋯+(k−2)+k
As k is even, you can say r = 2k and so:
S=2(1+2+3+⋯+(r−1)+r)
Now the sum of the first r numbers is well-known, it's the rth triangle number and we have:
1+2+3+⋯+(r−1)+r=r(r+1)/2
Now we can keep it simple and say 2k=4r and so:
S=2(1+2+3+⋯+(r−1)+r)=4r=2r(r+1)2=r(r+1)
So you can build a quadratic in r and so get k.
Let S be the sum of the even integers between 1 and k inclusive.
Then:
S=2+4+6+⋯+(k−2)+k
As k is even, you can say r = 2k and so:
S=2(1+2+3+⋯+(r−1)+r)
Now the sum of the first r numbers is well-known, it's the rth triangle number and we have:
1+2+3+⋯+(r−1)+r=r(r+1)/2
Now we can keep it simple and say 2k=4r and so:
S=2(1+2+3+⋯+(r−1)+r)=4r=2r(r+1)2=r(r+1)
So you can build a quadratic in r and so get k.
Answer:
K = 6
Step-by-step explanation:
Even integers between 1 and k will be 2, 4, 6, 8, 10.........(k-2), k
Therefore, number of even terms between 1 and k inclusive will be [tex]\frac{k}{2}[/tex]
Now we know these even integers will form an arithmetic sequence which has it's first term as 2 and a common difference of 2.
We know sum of an arithmetic sequence is represented by
[tex]S_{n}[/tex] = [tex]\frac{n}{2}[2a + (n -1)d][/tex]
where a = first term
n = number of therms
d = common difference
2k = [tex]\frac{k}{4}[2(2)+(\frac{k}{2}-1)2][/tex]
2k = [tex]\frac{k}{4}[4+k-2][/tex]
2k = [tex]\frac{k}{4}[k+2][/tex]
8k = k[k + 2]
8 = k + 2
k = 8 - 2
k = 6
