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It turns out that there are 79 individual even numbers and the average of each is 80. Hence the 79*80 = the sum of even numbers from 1 to n. Why is this true?

Lets use a smaller sum to illustrate the rule:

lets say the sum of even integers from 1 to n is 10*11 where n is an odd number. What does it mean "Where n is an odd number?" It means that when we find the value of the highest even number in the progression, we must add 1 to get n. That's all.

Back to 1 to n = 10*11. It's easier to see here becuase you can write out 10 * 11 = 110 2 + 4 + 6 + 8 + 10 + 12 + 14 + 16 + 18 + 20 + 22 + 24 + 26 + 28 + 30 + 32...that should cover it.

You won't need to do this on the exam once you see the rule. These steps are just to illustrate the rule.

If you have the first 5 even numbers, which is 2, 4 , 6, 8 and 10. The sum is 30. (2+8=10)(4+6=10) and (10) = sum of 30. what's the average of this? Average = 6. 30 / 5 numbers = 6. If you represented the total as a product of 2 numbers, it could be 5*6. Happens to be 5 numbers * the average of those numbers.

So 79*80 is the folowing. 79 total numbers, and 80 is the average of those numbers. It happens to be here that there are 79 even numbers. If you get a question like this again, the number of numbers will be the smaller number...79 rather than 80. So, now we double 79 to get the actual value of the highest even number because 79 is the number of even numbers. so 79 * 2 = 158. But n is an odd number as indicated in the stem, so 158 + 1 to get the next odd number that would not include another even number. 161 would include 160 and throw the total off.

What happens if you get something like this:

The sum of all even numbers from 1 to n, where n is odd, is 6320. They don't tell you it's 79*80. Are we screwed now? No.

Apply this rule:

You need to find 2 perfect squares. The nearest one that is below your total and the nearest one that is above your total. For the 79*80 example..79*79 is going to be 6241 and 80*80 is going to be 6400. These surround the value presented of 6320. What signifiance does this have?

If you take the square of the high number and subtract out the value given (6400 - 6320) you get 80.

If you take the number given and subtract out the low number, you get 79. (6320 - 6241). Magic!!!

These are your numbers you need to answer the question. It gives you 79 numbers at the average of 80. so 79*2 because it's even, then if the question says n is the next odd, the answer is 159.

Check this on a random example:

The sum of the even numbers between 1 and n is 39800 where n is odd. What is the value of n?

This is a crazy high number, but i've made it somewhat easy for you as you can see that 40,000 is higher and 4 is a square of 2, but that 4 has four zeros after it, so add half of those zeros to 2. So you have 200 (two zeros). You have to account for all zeros, but it's a square, so the other number would need the other "half" of the zeros in the original number. Square root of 16,000,000 is 4000. 6 zeros after 16 and 16 is the square of 4, so 4 + 3 zeros. Back to the task at hand.

So once you see that 200 is the closest perfect square to the high side, you just take 200 -1 to get 199 * 200. If you want, do the math and 200 * 199 = 39800. Follow the same rules as before. Double 199 to 398, add 1 if you need to know n as an odd number for n = 399.

If you get an odd progression, the total of all odd numbers from 1 to n...the total will always be a perfect square of (n+1)/2. For example:

If the sum of a certain number of odd integers is 441, how many integers are there? The answer will be \(sqrt{441}=21\)
_________________

------------------------------------ J Allen Morris **I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.

Can I ask someone to take a look at this one and provide a short explanation? Thank you.

The sum of the even numbers between 1 and n is 79*80, where k is an odd number, then n=? (A) 79 (B) 80 (C) 81 (D) 157 (E) 159

\(# \ of \ multiples \ of \ x \ in \ the \ range = \frac{Last \ multiple \ of \ x \ in \ the \ range \ - \ First \ multiple \ of \ x \ in \ the \ range}{x}+1\) (check this: totally-basic-94862.html and this: math-number-theory-88376.html).

Now, as as \(n\) is odd then the last even number between 1 and \(n\) will be \(n-1\) and the first even number will be 2. So, there are \(\frac{(n-1)-2}{2}+1=\frac{n-1}{2}\) even numbers (multiples of 2) between 1 and \(n\).

Next, the sum of the elements in any evenly spaced set is given by: \(Sum=\frac{first+last}{2}*# \ of \ terms\), the mean multiplied by the number of terms --> \(\frac{2+(n-1)}{2}*\frac{n-1}{2}=79*80\) --> \((n-1)(n+1)=158*160\) --> \(n=159\)

The sum of the even numbers between 1 and n is 79*80, where n is an odd number, then n=?

Alternatively, Sum of x even integers is x(x+1) = 79*80. So x = 79 i.e. there are 79 even integers. These 79 even integers lie between 1 and n. 2 will be first such even integer, next will be 4, next will be 6 and so on till we reach the last even integer 79*2 = 158. So all even integers from 2 to 158 lie between 1 and 159. So n must be 159. n cannot be 161/163/165... because then, more even integers will lie between 1 and n.

A Quick Recap:

- Sum of positive consecutive integers starting from 1 is \(\frac{x(x + 1)}{2}\) where x is the number of integers. 1+2+3+4+5+...+10 = 10*11/2

- Sum of positive consecutive even integers starting from 2 is \(x*(x + 1)\)where x is the number of even integers. 2+4+6+8+10 = 5*6 = 30

The explanation is simple. 2+4+6+8+10 = 2(1+2+3+4+5) By taking 2 common Sum = 2(5*6/2) = 5*6

- Sum of positive odd integers starting from 1 is \(x^2\) where x is the number of odd integers. 1+3+5+7+9+11 = \(6^2\)

I can derive it in the following way: Say x = 6. \(1+3+5+7+9+11 = (1+1)+(3+1)+(5+1)+(7+1)+(9+1)+(11 +1) - 6\) I add and subtract x from the right side. The right side becomes: \(2+4+6+8+10+12 - x = x(x+1) - x = x^2\)
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Re: The sum of the even numbers between 1 and n is 79*80, where [#permalink]

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17 Dec 2011, 04:18

Using the the Sum formula for Arithmetic Progression: n/2[2a+(n-1)d] a= first term=2 d= difference= 2 i get, n^2+n=79*80 n*(n+1)=79*80 so, n=79 What am i doing wrong?

Re: The sum of the even numbers between 1 and n is 79*80, where [#permalink]

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22 Mar 2012, 00:36

1

This post received KUDOS

Dear BDsunDevil,

One important thing to remember while deploying the formula s= n/2{2a+(n-1)d} is that n is the number of terms and not the last term, which a lot of students would wrongly assume while doing such problem. In that case the expression of sum would boil down to : s=n(n+1)=6320=79*80=> n=79, which is wrong! See, if you’re calculating the sum of even numbers from 1 to 7, the sum is 2+4+6=12 for 3 terms, In this example n=3? How, simple, n=(7-1)/2; On the similar lines the formula is all same for except the value of N which is now (n-1)/2; putting this value would get you: S=1/2*(n-1)/2*[2*2+{(n-1)/2 – 1}*2]=(n-1)(n+1)/4 As per the question, S=79*80=> (n-1)(n+1)=79*80*4=158*160=>n=159

- Sum of positive consecutive integers starting from 1 is \(\frac{x(x + 1)}{2}\) where x is the number of integers. 1+2+3+4+5+...+10 = 10*11/2

- Sum of positive consecutive even integers starting from 2 is \(x*(x + 1)\)where x is the number of even integers. 2+4+6+8+10 = 5*6 = 30

The explanation is simple. 2+4+6+8+10 = 2(1+2+3+4+5) By taking 2 common Sum = 2(5*6/2) = 5*6

- Sum of positive odd integers starting from 1 is \(x^2\) where x is the number of odd integers. 1+3+5+7+9+11 = \(6^2\)

I can derive it in the following way: Say x = 6. \(1+3+5+7+9+11 = (1+1)+(3+1)+(5+1)+(7+1)+(9+1)+(11 +1) - 6\) I add and subtract x from the right side. The right side becomes: \(2+4+6+8+10+12 - x = x(x+1) - x = x^2\)

Coming back to this question:

Sum of x even integers is x(x+1) = 79*80. So x = 79 i.e. there are 79 even integers. These 79 even integers lie between 1 and n. 2 will be first such even integer, next will be 4, next will be 6 and so on till we reach the last even integer 79*2 = 158. So all even integers from 2 to 158 lie between 1 and 159. So n must be 159. n cannot be 161/163/165... because then, more even integers will lie between 1 and n.
_________________

- Sum of positive consecutive integers starting from 1 is \(\frac{x(x + 1)}{2}\) where x is the number of integers. 1+2+3+4+5+...+10 = 10*11/2

- Sum of positive consecutive even integers starting from 2 is \(x*(x + 1)\)where x is the number of even integers. 2+4+6+8+10 = 5*6 = 30

The explanation is simple. 2+4+6+8+10 = 2(1+2+3+4+5) By taking 2 common Sum = 2(5*6/2) = 5*6

- Sum of positive odd integers starting from 1 is \(x^2\) where x is the number of odd integers. 1+3+5+7+9+11 = \(6^2\)

I can derive it in the following way: Say x = 6. \(1+3+5+7+9+11 = (1+1)+(3+1)+(5+1)+(7+1)+(9+1)+(11 +1) - 6\) I add and subtract x from the right side. The right side becomes: \(2+4+6+8+10+12 - x = x(x+1) - x = x^2\)

Coming back to this question:

Sum of x even integers is x(x+1) = 79*80. So x = 79 i.e. there are 79 even integers. These 79 even integers lie between 1 and n. 2 will be first such even integer, next will be 4, next will be 6 and so on till we reach the last even integer 79*2 = 158. So all even integers from 2 to 158 lie between 1 and 159. So n must be 159. n cannot be 161/163/165... because then, more even integers will lie between 1 and n.

So if we have to take odd consecutive odd integers we can directly square n?
_________________

So if we have to take odd consecutive odd integers we can directly square n?

Yes, sum of FIRST n consecutive positive odd integers = n^2 e.g. Sum of first 2 consecutive positive odd integers = 2^2 = 4 = 1 + 3 (first two consecutive odd integers)
_________________

Re: The sum of the even numbers between 1 and n is 79*80, where [#permalink]

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12 Nov 2013, 20:38

I solved by thinking about how you find the sum of a series of numbers; average of that set multiplied by how many numbers are in the set. Noting that it was only even numbers in the set A-C could be eliminated since their sums would be FAR too small. From there I thought about how many even numbers would be in the set if k was 157; in this case 78, with 156 being the final number in the set. Since 2 was the first, and 156 the last, the average would be 79. So this makes no sense for an answer.

Next I looked at 159. I see that there are 79 even integers between 0 and 159, ranging from 2-158. The average would be 80, leading to a sum of 80*79, thus E must be the answer

Re: The sum of the even numbers between 1 and n is 79*80, where [#permalink]

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25 Jun 2014, 23:22

VeritasPrepKarishma wrote:

chan4312 wrote:

The sum of the even numbers between 1 and n is 79*80, where n is an odd number, then n=?

Alternatively, Sum of x even integers is x(x+1) = 79*80. So x = 79 i.e. there are 79 even integers. These 79 even integers lie between 1 and n. 2 will be first such even integer, next will be 4, next will be 6 and so on till we reach the last even integer 79*2 = 158. So all even integers from 2 to 158 lie between 1 and 159. So n must be 159. n cannot be 161/163/165... because then, more even integers will lie between 1 and n.

A Quick Recap:

- Sum of positive consecutive integers starting from 1 is \(\frac{x(x + 1)}{2}\) where x is the number of integers. 1+2+3+4+5+...+10 = 10*11/2

- Sum of positive consecutive even integers starting from 2 is \(x*(x + 1)\)where x is the number of even integers. 2+4+6+8+10 = 5*6 = 30

The explanation is simple. 2+4+6+8+10 = 2(1+2+3+4+5) By taking 2 common Sum = 2(5*6/2) = 5*6

- Sum of positive odd integers starting from 1 is \(x^2\) where x is the number of odd integers. 1+3+5+7+9+11 = \(6^2\)

I can derive it in the following way: Say x = 6. \(1+3+5+7+9+11 = (1+1)+(3+1)+(5+1)+(7+1)+(9+1)+(11 +1) - 6\) I add and subtract x from the right side. The right side becomes: \(2+4+6+8+10+12 - x = x(x+1) - x = x^2\)

What I did was: (n-1)/2 is the number of terms. 2a = 4, (n-1)*2 = (n-1)*d

putting it all together: (n-1)*(4+(n-1)*2)/2 -> (n-1)(n+1) = 79*80. It would have been fine, except somewhere along the way I was supposed to multiply 79*80 by 2. I can't find what went wrong though....

What I did was: (n-1)/2 is the number of terms. 2a = 4, (n-1)*2 = (n-1)*d

putting it all together: (n-1)*(4+(n-1)*2)/2 -> (n-1)(n+1) = 79*80. It would have been fine, except somewhere along the way I was supposed to multiply 79*80 by 2. I can't find what went wrong though....

You have: 2 + 4 + ...+ (n-1) = 79*80

How many numbers are these? (n-1)/2 Note that everywhere in the formula where you have n, you will need to put (n-1)/2 because in the formula number of numbers is n.

Formula: Sum = n/2[2a + (n-1)*d] (n-1)/4 * [4 + {(n-1)/2 - 1}*2] = 79*80 On simplifying, you get (n-1)*(n+1) = 79*2*80*2 = 158*160