Find an Expression for a^n Where N Is an Arbitrary Positive Integer

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For any positive integer n, the sum of the first n positive integers [#permalink]   Updated on: 10 Jun 2021, 06:04

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For any positive integer n, the sum of the first n positive integers equals n(n+1)/2. What is the sum of all the even integers between 99 and 301?

A. 10,100
B. 20,200
C. 22,650
D. 40,200
E. 45,150

Originally posted by CharmWithSubstance on 11 May 2010, 14:57.
Last edited by Bunuel on 10 Jun 2021, 06:04, edited 2 times in total.

Updated.

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Re: For any positive integer n, the sum of the first n positive integers [#permalink]   19 Feb 2012, 22:11

For any positive integer n, the sum of the first n positive integers equals n(n+1)/2. What is the sum of all the even integers between 99 and 301?

A. 10,100
B. 20,200
C. 22,650
D. 40,200
E. 45,150

Approach #1:

Even integer between 99 and 301 represent evenly spaced set (aka arithmetic progression): 100, 102, 104, ..., 300. Now, the sum of the elements in any evenly spaced set is the mean (average) multiplied by the number of terms. (Check Number Theory chapter of Math Book for more:

http://gmatclub.com/forum/math-number-theory-88376.html

)

Average of the set: (largest+smallest)/2=(300+100)/2=200;
# of terms: (largest-smallest)/2+1=(300-100)/2+1=101 (check this:

http://gmatclub.com/forum/totally-basic ... ml#p730075

);

The sum = 200*101= 20,200.

Answer: B.

Approach #2:

Using the formula of the sum of the first n positive integers: n(n+1)/2.

100+102+...+300=2(50+51+..+150). Now, the sum of the integers from 50 to 150, inclusive equals to the sum of the integers from 1 to 150, inclusive minus the sum of the integers from 1 to 49, inclusive. 2(50+51+..+150)=2*(150(150+1)/2-49(49+1)/2)=20,200.

Answer: B.

Hope it helps.
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Re: For any positive integer n, the sum of the first n positive integers [#permalink]   27 Apr 2016, 07:42

gwiz87 wrote:

For any positive integer n, the sum of the first n positive integers equals n(n+1)/2. What is the sum of all the even integers between 99 and 301?

A. 10,100
B. 20,200
C. 22,650
D. 40,200
E. 45,150

Hi,

Hoping you can explain this one to me.

Although a formula is provided in this problem, we can easily solve it using a different formula:

sum = (average)(quantity)

Let's first determine the average.

In any set of numbers in an arithmetic sequence, we can determine the average using the formula:

(1st number in set + last number in set)/2

Remember, we must average the first even integer in the set and the last even integer in the set. So we have:

(100 + 300)/2 = 400/2 = 200

Next we have to determine the quantity. Once again, we include the first even integer in the set and the last even integer in the set. Thus, we are actually determining the quantity of even consecutive even integers from 100 to 300, inclusive.

Two key points to recognize:

1) Because we are determining the number of "even integers" in the set, we must divide by 2 after subtracting our quantities.

2) Because we are counting the consecutive even integers from 100 to 300, inclusive, we must "add 1" after doing the subtraction.

quantity = (300 – 100)/2 + 1

quantity = 200/2 + 1

quantity = 101

Finally, we can determine the sum.

sum = 200 x 101

sum = 20,200

Note that the reason this is easier than using the formula provided, is that the given formula would have to be applied several times since we don't want the total of all of the first 300 numbers. We'd have to remember to subtract the sum of the first 99 and divide by two to count only the even numbers. But we'd also have to account for the fact that the first and the last number in the set are both even. So even though a formula is given, it isn't very easy to use.

Answer is B.
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Re: For any positive integer n, the sum of the first n positive integers [#permalink]   19 Dec 2020, 15:31

Video solution from Quant Reasoning:
Subscribe for more:

https://www.youtube.com/QuantReasoning? ... irmation=1

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Re: For any positive integer n, the sum of the first n positive integers [#permalink]   25 Feb 2013, 07:42

Hi neerajeai, please note that the n(n+1)/2 shortcut formula is only applicable if the starting point is 1. Anytime you want to find the number of terms between two given numbers you should use the general formula ((first - last) / frequency) + 1. You can also multiply by the average at the end to get the sum.

In your case it is (((301-99)/2) + 1) * 200 = 102 * 200 = 20400. Answer choice C with a presumed typo in the unit digit?

Hope this helps!
-Ron
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Re: For any positive integer n, the sum of the first n positive integers [#permalink]   19 Apr 2018, 12:46

gwiz87 wrote:

For any positive integer n, the sum of the first n positive integers equals n(n+1)/2. What is the sum of all the even integers between 99 and 301?

A. 10,100
B. 20,200
C. 22,650
D. 40,200
E. 45,150

Here's one approach.

We want 100+102+104+....298+300
This equals 2(50+51+52+...+149+150)
From here, a quick way is to evaluate this is to first recognize that there are 101 integers from 50 to 150 inclusive (150 - 50 + 1 = 101)

To evaluate 2(50+51+52+...+149+150), let's add values in pairs:

....50 + 51 + 52 +...+ 149 + 150
+150+ 149+ 148+...+ 51 + 50
...200+ 200+ 200+...+ 200 + 200

How many 200's do we have in the new sum? There are 101 altogether.
101 x 200 = 20,200

Answer: B

Approach #2:

From my last post, we can see that we have 101 even integers from 100 to 300 inclusive.

Since the values in the set are equally spaced, the average (mean) of the 101 numbers = (first number + last number)/2 = (100 + 300)/2 = 400/2 = 200

So, we have 101 integers, whose average value is 200.
So, the sum of all 101 integers = (101)(200)
= 20,200
= B

Approach #3:

Take 100+102+104+ ...+298+300 and factor out the 2 to get 2(50+51+52+...+149+150)
From here, we'll evaluate the sum 50+51+52+...+149+150, and then double it.

Important: notice that 50+51+.....149+150 = (sum of 1 to 150) - (sum of 1 to 49)

Now we use the given formula:
sum of 1 to 150 = 150(151)/2 = 11,325
sum of 1 to 49 = 49(50)/2 = 1,225

So, sum of 50 to 150 = 11,325 - 1,225 = 10,100

So, 2(50+51+52+...+149+150) = 2(10,100) = 20,200

Answer: B

Cheers,
Brent
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Re: For any positive integer n, the sum of the first n positive integers [#permalink]   10 Jun 2021, 04:30

Sum of first n positive integers = \(\frac{n(n+1) }{ 2}\).

Even integers between 99 and 301 belong to the set { 100, 102, 104, 106….. 300}.
Therefore, sum of even integers between 99 and 301 = 100 + 102 + 104 + 106 + ………. + 300

Factoring out 2, we can write the expression as 2 (50 + 51 + 52 + ……. + 150). The term inside the bracket represents a sequence of positive integers. But, we cannot apply the given formula directly since these do not represent the FIRST n positive integers.

Therefore, (50 + 51 + 52 +…. 150) = (1+ 2 + 3 + … + 150) – (1+2+3+…. 49)

Value of the first bracket = \(\frac{150 * 151 }{ 2}\)

Value of the second bracket = \(\frac{49 * 50 }{ 2}\).

Therefore, (50 + 51 + 52 +…. 150) = \(\frac{150 * 151 }{ 2} - \frac{49 * 50 }{ 2}\) = \(\frac{50}{2}\) (3*151 – 49) = 25 (453 – 49) = 25 * 404.

This should be a small value over 20000 since 25 * 400 = 20000. The only value that fits is 20,200.

The correct answer option is B.

Hope that helps!
Aravind B T
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Re: For any positive integer n, the sum of the first n positive integers [#permalink]   19 Feb 2012, 18:36

For any positive integer n, the sum of the first n positive integers equals n(n+1)/2. What is the sum of all the even integers between 99 and 301?

a = first term = 100
l = last term = 300

Total number of terms = ((300-100)/2 ) + 1 = 101

sum = n (a + l ) / 2 = 101 (100 + 300 ) / 2 = 20200

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Re: For any positive integer n, the sum of the first n positive integers [#permalink]   26 Aug 2012, 09:30

In this question, they give you a formula to reference: the sum of the first n positive integers is n(n+1)/2

But wait, the question is about the sum of EVEN integers, not all positive integers.

We can understand the sum of even integers between 99 and 301 by somehow rewriting in terms of the expression they gave us. They were using the terminology of the first n positive integers.

OK, well sum of even integers between 99 and 301 is the same as between 100 and 300 - inclusive.

Same as the sum of the even integers up to 300 - sum of the even integers up to 98
Same as the sum of the first 150 even integers - sum of the first 49 even integers

So how do we translate this to the expression we were given? We need to translate our "even" integer expression into "positive integers" expression with our formula.

Sum of first 150 even integers = Sum of first 150 POSITIVE integers * 2
1, 2, 3, 4, 5 ===> 2, 4, 6, 8, 10 etc

Sum of first 49 even integers = Sum of first 49 POSITIVE integers * 2

2 * (n(n+1)/2) - 2 * (n(n+1)/2)

Where n=150 in the first case, and n = 49 in the second case

= 2* (150(151)/2) - 2 * (49(50)/2)
= 150*151 - 49*50
= 50 (3*151 - 49)
= 50 (453 - 49)
= 50 (404)
= 50 (400 + 4)
= 20,000 + 200
= 20,200

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Re: For any positive integer n, the sum of the first n positive integers [#permalink]   15 Nov 2018, 22:46

We are given the formula:

\(∑_1^n \frac{n(n+1)}{2}\)

We're asked to find sum of all the even numbers between \(99\) and \(301\)....

\(100+102+104+...+296+298+300\)

We can factor out \(2\) from this sum:

\(2 \times (50+51+52+...+148+149+150)\)

In the brackets we have sum of numbers from \(50\) to \(150\). We can use the given formula for the sum of \(1\) to \(150\) and subtract the sum of \(1\) to \(49\) to find the total of this sum.

\(2 \times (\frac{150 \times 151}{2}-\frac{49 \times 50}{2})\)

\(=(150)(151) – (49)(50)\)

\(=22,650 – 2,450\)

\(=20,200\)

The final answer is

.

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Re: For any positive integer n, the sum of the first n positive integers [#permalink]   11 May 2010, 15:54

CharmWithSubstance wrote:

For any positive integer n, the sum of the first n positive integers equals n(n+1)/2. What is the sum of all the even integers between 99 and 301?

A. 10,100
B. 20,200
C. 22,650
D. 40,200
E. 45,150

In any set of consecutive integers the sum = average*number of integers

average = 200
total = 150-50 + 1 = 101

150 because 300 is the 150th even integer
50 because 100 is the 50th even integer

101*200 = 20200
Therefore there are 101 integers

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Re: For any positive integer n, the sum of the first n positive integers [#permalink]   11 May 2010, 19:26

CharmWithSubstance wrote:

For any positive integer n, the sum of the first n positive integers equals n(n+1)/2. What is the sum of all the even integers between 99 and 301?

A. 10,100
B. 20,200
C. 22,650
D. 40,200
E. 45,150

sum of all even integer 1 to 301 = 2 * (1+2+...+150) = (2*150*151)/2 = 150*151
sum of all even integer 1 to 99 = 2 * (1+2+....+49) = (2*49*50)/2 = 49*50

required sum = 150*151 - 49*50 = 50*(453 - 49) =
404 * 50 = 20200

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Re: For any positive integer n, the sum of the first n positive integers [#permalink]   10 Jul 2010, 17:14

They are basically asking for the sum of even integers between/inclusive of 100 and 300.

Sum = \(100+102+ .... 300 = 2(50+51+52+ ....150)\) - We've converted even integers into consecutive integers for using this formula.

Now, let's just add and subtract 1 to 50 to this series, so we get:

\(2[(1+2+3 ..... 50+51+ ..... 150)-(1+2+3 ...50)]\)

So, we've broken this sum down into two sums of 1 .... x

Sum of 1 .... 150 = \(\frac{150(150+1)}{2} = 75*151\)

Similarly, sum of 1 .... 49 = \(\frac{49(49+1)}{2} = 25*49\)

So we get the answer to be: \(2[(75*151) - (25*49) = 20200\)

The reason why we do this is because in the sum of even integers, when 2 is factored out, you get consecutive integers as I've shown here. And since we have a formula for the first "n" integers and not for some random range of integers we try to reduce what we're asked to find to this form

Hope this answers your question.

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Re: For any positive integer n, the sum of the first n positive integers [#permalink]   11 Jul 2010, 05:36

thanks!! it makes sense now...what about if the problem was changed to ask for the sum of odd integers? can we amend the formula to get it? thanks!!

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Re: For any positive integer n, the sum of the first n positive integers [#permalink]   22 Oct 2010, 09:31

even integers between 99 and 301 are 100, 102, 104.....,300

for an arithmatic progression,a1,a2,a3....,an the sum is n* (a1+an)/2

so here.. n = 101, hence sum = 101 * 400/2 = 101 * 200 = 20200 : B

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Re: For any positive integer n, the sum of the first n positive integers [#permalink]   22 Oct 2010, 15:03

metallicafan wrote:

For any positive integer n, the sum of the first n positive integers equals \((n*(n+1))/2\). What is the sum of all the even integers between 99 and 301?
(A) 10,100
(B) 20,200
(C) 22,650
(D) 40,200
(E) 45,150

Sum = 100+102+...+300
= 100*101 + (0+2+4+...+200) [Taking 100 from each term in series, there are 101 terms]
= 100*101 + 2*(1+2+..+100) [Taking 2 common]
= 100*101 + 2*100*101/2 [Using formula given]
= 2*100*101
= 20200

Answer is (b)
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Re: For any positive integer n, the sum of the first n positive integers [#permalink]   07 Dec 2010, 06:30

can someone explain me why do we subtract 49 and no 50 ?

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Re: For any positive integer n, the sum of the first n positive integers [#permalink]   07 Dec 2010, 18:28

tt11234 wrote:

thanks!! it makes sense now...what about if the problem was changed to ask for the sum of odd integers? can we amend the formula to get it? thanks!!

The sum of n consecutive odd integers starting from 1 is \(n^2\)

e.g. sum of 1 + 3 + 5 + 7 = \(4^2\)

GMAT doesn't expect you to know it but it is a standard formula and doesn't hurt to know.
It can be easily derived using n(n+1)/2 formula. Let's wait for a day or two to see if someone comes up with a simple method of doing it.

tomchris wrote:

can someone explain me why do we subtract 49 and no 50 ?

We need to find the sum
50 + 51 + 52 +...+ 150 (The sum we need includes 50)

It can be done in two ways:

1. Using n(n + 1)/2

1 + 2 + 3 + ...49 + 50 + 51 + ....150 = 150.151/2
Also, 1 + 2 + 3 + ....+ 49 = 49.50/2

To get the required sum, we just subtract second equation from first.
We get: 50 + 51 + ...150 = 150.151/2 - 49.50/2

2. Using average of AP concept.
Note 50, 51...150 is an arithmetic progression. (Common difference between adjacent terms.)
The average of this AP = (first term + last term)/2 = (50 + 150)/2 = 100
Sum of AP = Average * No of terms = 100 * 101
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Re: For any positive integer n, the sum of the first n positive integers [#permalink]   08 Dec 2010, 17:32

Ok. Let's quickly review sum of some series:

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

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

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

So 100 + 102 + 104 + 106 + ...+ 300 can be solved in multiple ways.

It has been solved above in the following manner:

100 + 102 + 104 + 106 + ...+ 300 = 2( 50 + 51 + 52 + 53 + ...+ 150) Take out 2 common and find the sum in brackets.

50 + 51 + 52 + 53 + ...+ 150
We know the sum of consecutive integers but only when they start from 1. So what do we do?

We find the sum of first 150 numbers and subtract the sum of first 49 numbers from it. That will give us the sum of numbers from 50 to 150. Note that we subtract 49 numbers because 50 is part of our series. We need it so we do not subtract it.

50 + 51 + 52 + 53 + ...+ 150 = 150*151/2 - 49*50/2

Then 100 + 102 + 104 + 106 + ...+ 300 = 2( 150*151/2 - 49*50/2) = 20200
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Re: For any positive integer n, the sum of the first n positive integers [#permalink]   01 Jan 2011, 20:33

There is an easier way (without using formula) :

number of even integers from 99 and 301 = (300-100)/2 + 1 = 101
Average = (100+300)/2 = 200

Sum of all even integers between 99 and 301 = 101 * 200 = 20,200

Re: For any positive integer n, the sum of the first n positive integers [#permalink]

01 Jan 2011, 20:33

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