If the sum of the first n positive integers is S, what is the sum of

Question

If the sum of the first n positive integers is S, what is the sum of the first n positive even integers, in terms of S ?

(A) S/2
(B) S
(C) 2S
(D) 2S + 2
(E) 4S

Bunuel · Accepted Answer

1+2+3+...+n=S 2+4+6+...+2n=2(1+2+3+...+n)=2S. Or you can simply take n=2 --> 1+2=3=S --> 2+4=6=2S. Answer: C. Generally: Sum of n first integers: 1+2+...+n=\frac{1+n}{2}*n Sum of n first odd numbers: a_1+a_2+...+a_n=1+3+...+a_n=n^2 , where a_n is the last, n_{th} term and given by: a_n=2n-1 . Given n=5 first odd integers, then their sum equals to 1+3+5+7+9=5^2=25 . Sum of n first positive even numbers: a_1+a_2+...+a_n=2+4+...+a_n =n(n+1) , where a_n is the last, n_{th} term and given by: a_n=2n . Given n=4 first positive even integers, then their sum equals to 2+4+6+8=4(4+1)=20 . For more check Number Theory chapter of Math Book: math-number-theory-88376.html

KarishmaB · Answer

I see that most people would solve it by taking some value of n... that's great... One suggestion - don't shy away from taking the lowest and simplest value. 
I read 'sum of the first n positive integers', I say 'n = 1'... First 1 positive integer is 1 so S = 1. First 1 positive even integer will be 2 which is equal to twice of S. Only one options works out here so you are done. (In some cases, more than one option could work out e.g. if  2S^2  were another option. In that case you might want to check for n = 2 as well)

gmat1220 · Answer

Let n = 6. 
S = n(n+1)/2 = 3 * 7 = 21
Sum of first 6 even = 2 + 4 + 6 + 8 + 10 + 12 = 42 = 2S

Hence C.

Spidy001 · Answer

Answer is C

Given sum of 1 2 3.....n = S

sum of 2 , 4 , 6 ....2n = 2  = 2S

144144 · Answer

hmm... lets say the series is

{1,2,3,4,5} then S = 15
the sum of the even numbers - 6.

what am i missing?

Bunuel · Answer

The sum of the first 5 positive integers is 1+2+3+4+5=15;
The sum of the first 5 positive even integers is 2+4+6+8+10=30=2*15 (so the second sum shouldn't be the sum of the even numbers from the first list as you did).

144144 · Answer

damn... got confused bc of the S.

thanks!

stonecold · Answer

Here is what i did in this one =>
Let n=2 => S=1+2=3
Sum of first even numbers => 2+4=6
Clearly -> Sum = 2*S
Hence C.

AbhimanyuDhar · Answer

Well, this one can be solved in less than 10 secs if we know some formulas

Sum of firs n natural numbers = n(n+1)/2

Sum of first n even natural numbers = n(n+1)

Sum of first n off natural numbers is n^2

ok 
 we are given that n(n+1)/2=S
that means n(n+1)=2S and we know that n(n+1) is sum of first n even natural numbers . And that's the answer. 
2S

jfranciscocuencag · Answer

Hello!

Could someone mind to explain to me where do we infer that n(n+1) must be divided by 2?

I am really confused about that.

Thank you so much!

dabaobao · Answer

Take n = 2

n +: 1,2 
Sum = 3 = S

n = even: 2, 4
Sum = 6 = 2S

ANSWER: C

jtresley25 · Answer

The general rule for the first n even integers is that it's sum is 2x the sum of the first n integers. See below for example proof:

Sum  = 21
Sum  = 42

The even sum of the first n integers is 2x the sum of the first n integers, so the answer is 2S (C).

BrentGMATPrepNow · Answer

Given: The sum of the first n positive integers is S 
So, we can write:   S = 1 + 2 + 3 + 4 + . . . + n

Question: What is the sum of the first n positive even integers, in terms of S ? 
Let E = the sum of the first n positive even integers
So, we can write:  E = 2 + 4 + 6 + 8 + . . . + 2n 
Factor out 2 to get:  E = 2( 1 + 2 + 3 + 4 + . . . + n )

Since   S = 1 + 2 + 3 + 4 + . . . + n  , we can substitute to get:  E = 2( S )

Answer: C

Nipunh · Answer

Sum of first n even integers = n(n+1)
given n(n+1)/2=Sn
=> n(n+1)= 2S

aymanesabbar · Answer

2 is an even factor in this case and S=even*even it will be even and S is odd well and odd sum*even number(2) will be even. So simple as this.

Let"s say 2+16+23=41*even number(S*2) will be 82 an even outcome same goes for an even sum .

srosh2000 · Answer

I did it this way:

We know mean/n = sum (S) -- (1)
Now, we are asked the sum of EVEN integers so the denominator becomes n/2 (half the numbers in the set are going to be even).

=> mean/(n/2) = S_even 
=> 2*mean/n = S_even
Plug in (1) in (2)

=> 2*S = S_even

So ans is (C).

theretheree4 · Answer

Can someone let me know is there any formula or rule that they are using because I can simply take a four values like 1234 and using the sum formula by n/2(first+last) I get 10 and if do for positive even that is 2468, I’ll get the 20 so I can directly say it’s twice. What am I missing here?

egmat · Answer

Hi theretheree4,

Good news first:  you're not missing anything.  Your method is completely valid, and it's the same plug-in approach several people used in this thread (Bunuel with n= 2 , gmat1220 with n= 6 ).

What you did:
- First  4  integers:  1+2+3+4 = 10 = S 
- First  4  even integers:  2+4+6+8 = 20 = 2S

That's a legitimate way to get to  C . There's no rule you overlooked - you basically  rediscovered  why the answer is  2S : every even term is just double its matching counting number ( 2  = 2· 1 ,  4  = 2· 2 , ...), so the whole sum doubles.

The one habit to add

When you test with a single number, there's a small risk worth knowing about: sometimes  more than one answer choice  produces your target value, and then a single case can't tell them apart. KarishmaB flagged exactly this earlier in the thread - she noted that if an option like  2S2  had been present, one value of n might not be enough to separate it from the right answer.

So the safe move is a quick scan: does  only one  choice hit your value? Using your S =  10 :
- A) S/ 2  =  5 
- B) S =  10 
- C)  2S  =  20  ✓
- D)  2S+2  =  22 
- E)  4S  =  40

Only  C  gives  20 , so your single test is airtight here. You're done.

Why the caution matters (quick illustration)

Watch how a small number can create a tie. Suppose a choice  2S2  had been on the list, and you'd picked the smallest case, n= 1 , so S =  1 :
-  2S  = 2· 1  =  2 
-  2S2  = 2· 1 2 =  2  - same value!

Now one case can't decide between them. Test a second case, n= 2 , so S =  3 :
-  2S  =  6 
-  2S2  = 2· 3 2 =  18  - now they split apart.

That's the whole reason to glance at the other choices (or try a second n): to make sure your number didn't accidentally match two answers. In  this  question it didn't, so your approach nailed it.

Answer: C

If the sum of the first n positive integers is S, what is the sum of

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If the sum of the first n positive integers is S, what is the sum of

Prep Toolkit

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