if the sum of the even integers between 1 and k, inclusive, is equal to 2k, what is the value of k?
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.
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 \(\frac{k}{2}\)
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
\(S_{n}\) = \(\frac{n}{2}[2a + (n -1)d]\)
where a = first term
n = number of therms
d = common difference
2k = \(\frac{k}{4}[2(2)+(\frac{k}{2}-1)2]\)
2k = \(\frac{k}{4}[4+k-2]\)
2k = \(\frac{k}{4}[k+2]\)
8k = k[k + 2]
8 = k + 2
k = 8 - 2
k = 6