If the sum of the first n positive integers is x, then which

Question

If the sum of the first n positive integers is x, then which of the following is the sum of the first 2n even positive integers?

(A) 2x + 2n^2
(B) 3x + 4n^2
(C) 4x + 2n
(D) 4x + n^2
(E) 4x + 2n^2

OE
 Use Picking Numbers
If n = 2, sum of the first 2 positive integers is (1 + 2) = 3. x = 3.
Sum of first 2n = 2(2) = 4 even integers is (2 + 4 + 6 + 8) = 20.
Plug in n=2, x=3 to answer choices to find the answer that is = 20
(A): 2x + 2n^2 = 2(3) + 2(2^2) = 6 + 2(4) = 6 + 8 = 14.Eliminate.
(B): 3x + 4n^2 = 3(3) + 4(2^2) = 9 + 4(4) = 9 + 16 = 25. Eliminate.
(C): 4x + 2n = 4(3) + 2(2) = 12 + 4 = 16. Eliminate.
(D): 4x + n^2 = 4(3) + 2^2 = 12 + 4 = 16. Eliminate.
(E): 4x + 2n^2 = 4(3) + 2(2^2) = 12 + 2(4) = (12 + 8) = 20 (= Answer)

KarishmaB · Accepted Answer

There are various ways in which you can arrive at the answer here. I prefer  number picking .

Say n = 1. Then x = 1.
Sum of 2n even numbers = 2+4 = 6
Put x = 1, n = 1 in the options. Only (C) and (E) give 6.
Now, say n = 2 and try this only on (C) and (E).

Another Method:

Given: 1+ 2+3+4..+n = x

Required: 2 + 4 + 6 + ...+ 2n +  (2n+2) + (2n+4) +....+ (2n+2n)  (total 2n terms)
Take 2 common: 2(1 + 2 + 3 +... n) + 2n*n + 2(1 + 2 + 3 + ... + n)
From the n highlighted terms, you collected all 2n to get 2n*n. You were left with 2 + 4 + 6 +...2n from which you again take 2 common.

= 2x + 2n*n + 2x
= 4x + 2n^2

akashganga · Answer

n(n+1)/2 = x
or n^2 + n = 2x (1)

If the series has first 2n numbers, sum of n odd numbers = n^2.
So sum of n even numbers= Sum of first 2n numbers - n^2

In this case we have 2n even numbers so 2n odd numbers will be there, hence a total of 4n first numbers.

Sum of first 4n numbers = 4n(4n+1)/2 -----------(2)

Sum of first 2n odd numbers is (2n)^2 = 4n^2 ---------------- (3)

So sum of 2n even numbers is (2)-(3)

or 4n^2+2n

or 2n^2 + 2 (n^2+n)

or 2n^2 + 2(2x)

or 2n^2 + 4x

(E)

DJ

goodyear2013 · Answer

n(n+1)/2 = x is a rule for sum of n positive integers, then.
Thanks. That helps!

goodyear2013 · Answer

Hi Karishma,

Could you complete from this point onward to choose (C) or (E), please.

KarishmaB · Answer

Say n = 2
Sum of first 2 positive integers = x = 1+2 = 3
Sum of first 2n (= 4) even positive integers = 2 + 4 + 6 + 8 = 20

We want to see which of (C) and (E) will give 20.

(C) 4x + 2n
4*3 + 2*2 = 16

(E) 4x + 2n^2
4*3 + 2*2^2 = 20

(E) is the answer

kunal555 · Answer

It is given that sum of first n positive integers is x
or x= n(n+1)/2
or 2x = n(n+1)............(1)

Sum of first n positive even integers is given by n(n+1)
But we are asked for the sum of first 2n positive integer. so the sum will be

2n(2n+1)
or 2n(n+1 +n)
or 2n(n+1) + 2n^2
or 2(2x) + 2n^2....................(putting the value of n(n+1) from 1)
or 4x+2n^2

Answer:-E

mvictor · Answer

what the hell the question asks? spent 3 minutes just trying to figure it out.
i understand that the sum of n numbers is smth, well, we know that n(n+1)/2=x (formula of the sum of n consecutive numbers)
but 2n - what is this? how to understand what it means?

VeritasPrepKarishma mentioned that, if we put n=1, then x is 1.
2n is thus 2, and it must have 2 elements 1,2 - that's it. where did the 4 come out? since the list contains only 1 and 2?

kunal555 · Answer

Sum of first n number = n(n+1)/2
Sum of first n odd numbers = n^2 
Sum of first n even numbers = n(n+1) ............(1)

Remember these formulas. They will help you.

Now, it is given that sum of first n natural numbers = x
or n(n+1)/2 =x 
or n(n+1) = 2x 
or n^2 + n = 2x ............(2)

We have to find the sum of first 2n natural numbers.
From (1) we know the sum of first n even numbers.
So the sum of first 2n even numbers will be
2n(2n+1)
or 2(2n^2+n)
or 2(n^2 + n^2 +n)
Putting the value of n^2+n from (2) in above expression, we have
2(n^2 + 2x)
or 2n^2+4x

hatemnag · Answer

Hi Karishma

Do formulas like N(N+1)/2 and 2n(2n+1) play a role in solving such questions.
please help.

KarishmaB · Answer

Yes you can solve using those but it will require some algebraic manipulations

n(n+1)/2 = x
n(n+1) = 2x

Sum of first 2n even integers:

2 + 4 + 6 + ...4n

= 2*(1 + 2 + 3 + ...2n)

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

= 2*n * (2n + 1)

= 2*n*

= 2*n*n + 2*n*(n + 1)

= 2n^2 + 2*2x

= 2n^2 + 4x

Answer (E)

a2menthiere · Answer

My answer a little bit more " theorical" but no way to get lost with

law258 · Answer

Say we have the following numbers:

n=4:
1 2 3 4

Sum = 10 = x

This means that 2n=8 & we will have the following numbers for the second part of the problem:

2 4 6 8 10 12 14 16

Sum = 72

Try manipulating the answers and you'll find that E works: 40+ 2(4^2) = 72

BrentGMATPrepNow · Answer

I think the  input-output approach  is probably the fastest here.

The sum of the first  n  positive integers is  x   
Let's see what happens when n =  3 
So, the sum of the first  3  positive integers = 1 + 2 + 3 =  6 
This means, when n =  3 , x = n =  6

Which of the following is the sum of the first 2n even positive integers?  
If n =  3 , then 2n =  6 
So, we must find the sum of the first  6  even numbers: 2 + 4 + 6 + 8 + 10 + 12 =   42  
In other words, when we input n =  3 , x = n =  6 , the answer to the question (i.e., the output) is   42  
Now we'll examine the five answer choices, and see which one yields an output of   42   when we input n =  3 , x =  6 
(A) 2( 6 ) + 2( 3 ²) =   40  . NO GOOD. We want   42  
(B) 3( 6 ) + 4( 3 ²) =   54  . NO GOOD. We want   42  
(C) 4( 6 ) + 2( 3 ) =   30  . NO GOOD. We want   42  
(D) 4( 6 ) +  3 ² =   33  . NO GOOD. We want   42  
(E) 4( 6 ) + 2( 3 ²) =   42   PERFECT!

Answer: E

Cheers, 
Brent

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