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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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New post 05 May 2017, 00:28
karishma Bunuel

The formula to find the sum of FIRST n nos. = n(n+1)/2

Set 1: 1,2,3,4...........10
n= 10 (no. of terms)

Sum (using the above formula)= 10*11/2= 55


Set 2: 11,12,13,14,...............20
n( again)= 10

Sum (using the above formula)= 10*11/2= 55

Which is wrong because the sum of set 2 is 155 and not 55.

My query is: This formula can be used only to find the sum of n nos. starting from 1 ?

That is why in the solution given by Bunuel:
As we have to find the sum of nos. from 50 to 150
Step 1: Sum of nos. from 1 to 150 (Sum 1)
Step 2: Sum of nos. from 1 to 49 (Sum 2)
Step 3: Subtract Sum 2 from sum 1 to get sum of nos. from 50 to 150

1,2,3,4........................49, 50, 51,...............150
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Re: For any positive integer n, the sum of the first n positive integers  [#permalink]

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New post 23 Jun 2017, 00:02
fameatop wrote:
This problem can be solved with an alternative formula:-
Sum of first 'n' even integers is given by - n(n+1), where n=2,4,6,......any even integer


I am not able to get the right answer using this formula.
To find n:
(300-100)/2 + 1 = 101.
According to the formula mentioned it is n*(n+1) = 101 * 102. I am not understanding where i am going wrong :(
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Re: For any positive integer n, the sum of the first n positive integers  [#permalink]

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New post 27 Jul 2017, 06:31
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Bang2919 wrote:
fameatop wrote:
This problem can be solved with an alternative formula:-
Sum of first 'n' even integers is given by - n(n+1), where n=2,4,6,......any even integer


I am not able to get the right answer using this formula.
To find n:
(300-100)/2 + 1 = 101.
According to the formula mentioned it is n*(n+1) = 101 * 102. I am not understanding where i am going wrong :(




You are missing a very important point here.
The question has mentioned that the sum of FIRST N POSITIVE INTEGERS EQUALS n(n+1)/2.

We have to just find the sum of even nos. from 100 to 300.

100,102,104,106,.....................,300

You have found correctly that the number of terms is 101.

But this formula can only be used for first n positive integers i.e. 1,2,3,4,5................................and so on.

Here the series is not starting from 1 but from 100.

Therefore

To use this formula, we will have to first find the sum of first 150 terms and then subtract the sum of first 49 terms.

Why so?

We have to find: 100+102+104+106+......................+300

This can also be written as:
2(50+51+.................+150)

We need the sum of 50th term+51st term, and so on till 150th term.

After you subtract the sum of 1st term till 49th term from the sum of 1st term till 150th term, multiply the answer by 2.

I hope you get it.


Otherwise there are some other great methods discussed here.

One more method could be:

nth term= a+(n-1)d
300= 100+(n-1)2
n= 101


Sum= n/2 (First term+Last term)
Sum= 101/2 *(100+300) = 20200


I hope it helps.
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Re: For any positive integer n, the sum of the first n positive integers  [#permalink]

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New post 28 Dec 2017, 19:06
Bunuel, niks18


Quote:
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.




This goes in to complex multiplication of a 2 digit with 3 digit, (even taking out common did not help)
Any easier way for such calculations?
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Re: For any positive integer n, the sum of the first n positive integers  [#permalink]

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New post 28 Dec 2017, 19:19
It is clear that to calculate the sum of even integers from 99 to 301:
a= 100; d=2;
To calculate the number of terms we know the last term = 300
100 + (n-1)*2 = 300 => n= 101

sum =(n/2)*(2a + (n-1)*d)
=(101/2)*(200+100*3) = 20200
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Re: For any positive integer n, the sum of the first n positive integers  [#permalink]

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New post 29 Dec 2017, 11:51
adkikani wrote:
Bunuel, niks18


Quote:
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.




This goes in to complex multiplication of a 2 digit with 3 digit, (even taking out common did not help)
Any easier way for such calculations?


Hi adkikani

you can use AP formula to find the number of even terms bewteen 99 & 301 and then use the sum formula for an AP series to calcualte the final sum

\(T_n=a_1+(n-1)d\), here \(T_n=300, a_1=100, d=2\), find \(n\)

then Sum, \(S_n=\frac{n(T_n+a_1)}{2}\).

the calculation should be simple.
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Re: For any positive integer n, the sum of the first n positive integers  [#permalink]

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New post 31 Dec 2017, 17:29
niks18

I was confused if the second formula was applied only when my sequence is starting from 1.
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Re: For any positive integer n, the sum of the first n positive integers  [#permalink]

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New post 31 Dec 2017, 21:19
adkikani wrote:
niks18

I was confused if the second formula was applied only when my sequence is starting from 1.


Hi adkikani

The formula is a simple derivation of AP series that starts from 1 and is consecutive

So the Sum of AP series is \(S_n=\frac{n[2a+(n-1)d]}{2}=\frac{n}{2}\)[First term + Last Term]

if the sequence starts from 1 and is consecutive so your First term=1 and Last Term=number of terms in the series (because sequence is consecutive)

so \(S_n=\frac{n(1+n)}{2}\)

Let's take a simple example. if I have to find sum of numbers from 6 to 10, then I can do that by first finding sum of all Numbers from 1 to 10 and then from this sum subtract sum of numbers from 1 to 5.

In the number line this can be represented as

1...2...3...4...5...6...7...8...9...10. Here First term =1 and last term=10=number of terms in the sequence

As you can clearly see to find sum of numbers from 6 to 10, we simply need to remove sum of numbers from 1 to 5 out of the total sum i.e from 1 to 10.

Now in this question we need to find sum of number from 50 to 150 so it will be Total sum of numbers from 1 to 150 less sum of numbers from 1 to 49
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Re: For any positive integer n, the sum of the first n positive integers  [#permalink]

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New post 19 Apr 2018, 13:46
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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

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

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New post 19 Apr 2018, 13:47
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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


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

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

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New post 19 Apr 2018, 13:48
Top Contributor
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


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]

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New post 20 Apr 2018, 01:57
Step 1: The formula is unnecesary and in my opinion is thrown there to put us offguard. We can solve this using the following two formulae:
Step 2: Find amount of integeres between 99 and 301. (First-Last/Increment+1) = 101
Step 3: Find the average of the set of numbers between 99 and 301. Since this is an evenly spaced set (constant increments of 2) We can find the mean of this number set by taking the average of the first and last terms. 99+301 = 400 / 2 = 200.
Step 4: Multiply the mean * number of integers in the set = 200*101 = 20,200.
Answer choice B.
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Re: For any positive integer n, the sum of the first n positive integers &nbs [#permalink] 20 Apr 2018, 01:57

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