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# For any positive integer n, the sum of the first n positive integers

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Re: For any positive integer n, the sum of the first n positive integers [#permalink]
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the sum of 1st even numbers = n(n+1)
the sum of first even numbers till 301 is 150*(150+1) =22650
the sum of first even numbers till 99 is 49*(49+1) =2450

22650-2450=20200

this method is not perfect. I wrote it just to show one more method
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Re: For any positive integer n, the sum of the first n positive integers [#permalink]
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Easier way would be using sum of all n even integers = n * (n+1)

Hence from 1-301 there are 150 even integers hence sum = 150 * 151 = 22650
From 1- 99 there are 49 even integers ( - 1 as 100 is not part of this) = 49 * 50 = 2450

So sum from 99-301 = 22650 -2450 = 20200
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Re: For any positive integer n, the sum of the first n positive integers [#permalink]
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This is another way to tackle this question:

It is given that the sum of the first n positive integers is n(n+1)/2. I didn't really use this formula.

I remembered this formula instead (and I think you should remember it too!):

The sum of the elements in any evenly spaced set is: (mean)*(# of terms)

Steps:
1. First, find the mean
The questions asked to find the sum of all the EVEN integers between 99 and 301. So that means we need to look at EVEN numbers only. So really the range we are looking at is between 100 and 300. So the mean of these evenly spaced numbers between 100 and 300 is 200. --> this is the mean (no calculation necessary here because it's an evenly spaced range so its very straight forward to find the mean; just get the middle number!)

2. Second, find the # of EVEN terms.
We stated in step 1 that the EVEN number range is between 100 and 300. So here we can utilise this formula:
[(last term -first term)/2] +1

which comes to:
[(300-100)]/2 + 1
= (200/2) +1
= 100+1
= 101 --> this is the # of EVEN terms in the range 100 and 300

Coming back to this: The sum of the elements in any evenly spaced set is: (mean)*(# of terms)

(mean)*(# of terms)
=200* 101
=20200

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Re: For any positive integer n, the sum of the first n positive integers [#permalink]
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Sum of all the even integers between 99 and 301:

The first even integer will be 100 and the last will be 300.

The common difference will be 2.

Number of terms: 300 = 100 + (n-1)2

=> 200 = 2n-2

=> 202 = 2n

=> n = 101

Sum: $$\frac{n}{2}$$[first term + last term]

=> $$\frac{101}{2}$$[100+300]

=> $$\frac{101}{2}$$ * 400

=> 101 *200

=> 20200

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Re: For any positive integer n, the sum of the first n positive integers [#permalink]
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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]
To find the sum of all the even integers between 99 and 301, we first need to determine the number of even integers in that range.

The largest even integer within the given range is 300, and the smallest even integer is 100. The set of even integers between 99 and 301 can be expressed as {100, 102, 104, ..., 298, 300}.To find the number of terms in this arithmetic sequence, we can use the formula for the nth term of an arithmetic sequence:

an = a1 + (n - 1)d

In this case, a1 = 100 (the first term), d = 2 (the common difference), and we need to find n.

300 = 100 + (n - 1)2
200 = 2n - 2
202 = 2n
n = 101

Therefore, there are 101 even integers between 99 and 301.

Now, we can calculate the sum using the formula for the sum of an arithmetic series:

Sn = (n/2)(a1 + an)
Sn = (101/2)(100 + 300)
Sn = (101/2)(400)
Sn = 101 * 200
Sn = 20,200

The sum of all the even integers between 99 and 301 is 20,200.

Therefore, the answer is option B.
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Re: For any positive integer n, the sum of the first n positive integers [#permalink]
Bunuel 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

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: https://gmatclub.com/forum/math-number- ... 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: https://gmatclub.com/forum/totally-basi ... ml#p730075);

The sum = 200*101= 20,200.

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.

Hope it helps.

In approach no. 1 - what would have been the difference if they wouldnt have mentiond 'even' numbers?
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Re: For any positive integer n, the sum of the first n positive integers [#permalink]
Rajeet123 wrote:
Bunuel 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

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: https://gmatclub.com/forum/math-number- ... 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: https://gmatclub.com/forum/totally-basi ... ml#p730075);

The sum = 200*101= 20,200.

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.

Hope it helps.

In approach no. 1 - what would have been the difference if they wouldnt have mentiond 'even' numbers?

If we were asked to find the sum of the integers between 99 and 301, inclusive, we can utilize the fact that the sum of the elements in any evenly spaced set is given by: $$Sum=\frac{a_1+a_n}{2}*n$$, the mean multiplied by the number of terms. Therefore, the answer would be (99 + 301)/2*203 = 40,600.
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Re: For any positive integer n, the sum of the first n positive integers [#permalink]
Given: For any positive integer n, the sum of the first n positive integers equals n(n+1)/2.
Asked: What is the sum of all the even integers between 99 and 301?

Sum of all even integers between 99 and 301 = 100 + 102 + .... + 300 = (2 + 4 + .... + 300) - (2 + 4 + ... + 98) =
= 2 (1+ 2+ .... + 150) - 2 (1+2+... + 49) = 150*151 - 49*50 = 50 (453 - 49) = 50*404 = 20200

IMO B
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Re: For any positive integer n, the sum of the first n positive integers [#permalink]
My method was similar to approach #1 here, please let me know if this is wrong

No. of terms = 300-100 + 1 = 201 terms

Sum of 201 terms in this sequence = (201(202)) / 2 = 20200
Bunuel 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

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: https://gmatclub.com/forum/math-number- ... 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: https://gmatclub.com/forum/totally-basi ... ml#p730075);

The sum = 200*101= 20,200.

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.

Hope it helps.

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