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Hitman42 [59]
3 years ago
15

If the sum of the even integers between 1 and k, inclusive, is equal to 2k, what is the value of k?

Mathematics
2 answers:
adelina 88 [10]3 years ago
6 0

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

lorasvet [3.4K]3 years ago
3 0
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:
<span><span>S=2+4+6+⋯+(k−2)+k</span></span>

As k is even, you can say r = 2k and so:

<span><span>S=2(1+2+3+⋯+(r−1)+r)</span></span>

<span>Now the sum of the first </span><span>r</span><span> numbers is well-known, it's the </span><span>r</span>th triangle number and we have:

<span><span>1+2+3+⋯+(r−1)+r=<span><span>r(r+1)/</span>2</span></span></span>
<span>Now we can keep it simple and say </span><span><span>2k=4r </span></span><span>and so:</span>

<span><span>S=2(1+2+3+⋯+(r−1)+r)=4r=2<span><span>r(r+1)</span>2</span>=r(r+1)</span></span>

<span>So you can build a quadratic in </span><span>r</span><span> and so get </span><span>k.</span>
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