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09 Jan 2014, 11:51
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
57% (02:46) correct
43%
(02:54)
wrong
based on 538
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History
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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
(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)
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)
09 Jan 2014, 22:12
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goodyear2013
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
(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)
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)
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
General Discussion
akashganga
SDA Bocconi Thread Master
Joined: 27 Dec 2012
Last visit: 18 Mar 2016
Posts: 26
Given Kudos: 34
Location: India
Concentration: Technology, Entrepreneurship
Schools: Bocconi '17 ISB '17 (WL) IIMA (II) Oxford'17 IE '18 IIMB (D)
GMAT 1: 660 Q48 V33
GMAT 2: 730 Q49 V40
WE:Engineering (Energy)
09 Jan 2014, 13:19
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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
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
