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# If J = 2 + 4 + 6 + 8 + . . . 98 + 100, and K = 1 + 3 + 5 + 7 + . . . +

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SVP
Joined: 11 Sep 2015
Posts: 2051
If J = 2 + 4 + 6 + 8 + . . . 98 + 100, and K = 1 + 3 + 5 + 7 + . . . + [#permalink]

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08 May 2017, 06:06
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Difficulty:

95% (hard)

Question Stats:

43% (02:12) correct 57% (01:41) wrong based on 150 sessions

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If J = 2 + 4 + 6 + 8 + . . . 98 + 100, and K = 1 + 3 + 5 + 7 + . . . + 97 + 99, then $$1^2 - 2^2 + 3^2 - 4^2 + 5^2 - 6^2 + . . . . . + 97^2 - 98^2 + 99^2 - 100^2 =$$

A) J² - K²
B) -50(J² - K²)
C) -K - J
D) K² - J²
E) (-J - K)²

*kudos for all correct solutions
[Reveal] Spoiler: OA

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Brent Hanneson – Founder of gmatprepnow.com

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Joined: 05 Mar 2015
Posts: 962
If J = 2 + 4 + 6 + 8 + . . . 98 + 100, and K = 1 + 3 + 5 + 7 + . . . + [#permalink]

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08 May 2017, 06:53
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GMATPrepNow wrote:
If J = 2 + 4 + 6 + 8 + . . . 98 + 100, and K = 1 + 3 + 5 + 7 + . . . + 97 + 99, then $$1^2 - 2^2 + 3^2 - 4^2 + 5^2 - 6^2 + . . . . . + 97^2 - 98^2 + 99^2 - 100^2 =$$

A) J² - K²
B) -50(J² - K²)
C) -K - J
D) K² - J²
E) (-J - K)²

*kudos for all correct solutions

simply put j= 2+4
k= 1+3

then 1^2- 2^2 + 3^2 - 4^2 = 1-4+9-16= -10

also j= 2+4= 6 && k = 1+3 =4
j^2= 36 && k^2 =16

just plug in values to option to get -10 as our answer

only option C does

Ans C
Math Expert
Joined: 02 Aug 2009
Posts: 5652
Re: If J = 2 + 4 + 6 + 8 + . . . 98 + 100, and K = 1 + 3 + 5 + 7 + . . . + [#permalink]

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08 May 2017, 07:25
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Expert's post
GMATPrepNow wrote:
If J = 2 + 4 + 6 + 8 + . . . 98 + 100, and K = 1 + 3 + 5 + 7 + . . . + 97 + 99, then $$1^2 - 2^2 + 3^2 - 4^2 + 5^2 - 6^2 + . . . . . + 97^2 - 98^2 + 99^2 - 100^2 =$$

A) J² - K²
B) -50(J² - K²)
C) -K - J
D) K² - J²
E) (-J - K)²

*kudos for all correct solutions

Hi,

$$1^2 - 2^2 + 3^2 - 4^2 + 5^2 - 6^2 + . . . . . + 97^2 - 98^2 + 99^2 - 100^2 =$$ .....
Here take all inpairs $$1^2-2^2$$, $$3^2-4^2$$, and so on till $$99^2-100^2$$..
1^2-2^2=(1-2)(1+2)=-1(1+2)=-1-2...
3^2-4^2=(3-4)(3+4)=-1(3+4)=-3-4..
So the equation becomes -1-2-3-4-......-99-100=-(1+2+3+4+...+99+100)= -[(2+4+6....+98+100)+(1+3+5+...+97+99)]=-[(j)+(k)]=-j-k
C
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Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

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Re: If J = 2 + 4 + 6 + 8 + . . . 98 + 100, and K = 1 + 3 + 5 + 7 + . . . + [#permalink]

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08 May 2017, 09:52
1
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GMATPrepNow wrote:
If J = 2 + 4 + 6 + 8 + . . . 98 + 100, and K = 1 + 3 + 5 + 7 + . . . + 97 + 99, then $$1^2 - 2^2 + 3^2 - 4^2 + 5^2 - 6^2 + . . . . . + 97^2 - 98^2 + 99^2 - 100^2 =$$

A) J² - K²
B) -50(J² - K²)
C) -K - J
D) K² - J²
E) (-J - K)²

*kudos for all correct solutions

We can break the problem into a^2 - b ^2 = 1^2 - 2^2 = (1+2) (1-2) = -3
similarly other pair will give = -7 ,next pair will give = -11
final pair will give -199
Now the question stem is reduced to below seq:
-3 -7-11.....-199

-3 = -(1+2)
-7 = -(3+4)
-11 = -(5+6)

-(1+3+5....)-(2+4+6...)
-(j)-(k)
-(j+k)..Ans
SVP
Joined: 11 Sep 2015
Posts: 2051
Re: If J = 2 + 4 + 6 + 8 + . . . 98 + 100, and K = 1 + 3 + 5 + 7 + . . . + [#permalink]

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10 May 2017, 13:07
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Expert's post
Top Contributor
2
This post was
BOOKMARKED
GMATPrepNow wrote:
If J = 2 + 4 + 6 + 8 + . . . 98 + 100, and K = 1 + 3 + 5 + 7 + . . . + 97 + 99, then $$1^2 - 2^2 + 3^2 - 4^2 + 5^2 - 6^2 + . . . . . + 97^2 - 98^2 + 99^2 - 100^2 =$$

A) J² - K²
B) -50(J² - K²)
C) -K - J
D) K² - J²
E) (-J - K)²

We have several differences of squares hiding in the expression 1² - 2² + 3² - 4² + 5² - 6² + . . . . . + 97² - 98² + 99² - 100²

1² - 2² + 3² - 4² + 5² - 6² + . . . . . + 97² - 98² + 99² - 100² = 1² - 2² + 3² - 4² + 5² - 6² + . . . . . + 97² - 98² + 99² - 100²
= (1 - 2)(1 + 2) + (3 - 4)(3 + 4) + (5 - 6)(5 + 6) + . . . . . + (97 - 98)(97 + 98) + (99 - 100)(99 + 100)
= (-1)(1 + 2) + (-1)(3 + 4) + (-1)(5 + 6) + . . . . . + (-1)(97 + 98) + (-1)(99 + 100)
= (-1)[(1 + 2) + (3 + 4) + (5 + 6) + . . . . . + (97 + 98) + (99 + 100)]
= (-1)(1 + 2 + 3 + 4 + . . . . . 97 + 98 + 99 + 100)

IMPORTANT: within the sum, 1 + 2 + 3 + 4 + . . . . . 97 + 98 + 99 + 100, we have all of the ODD integers from 1 to 99 inclusive, and we have all of the EVEN integers from 2 to 100 inclusive.

So, we can say that 1 + 2 + 3 + 4 + . . . . . 97 + 98 + 99 + 100 = K + J

So, we're replace 1 + 2 + 3 + 4 + . . . . . 97 + 98 + 99 + 100 with K + J.
We get: (-1)(1 + 2 + 3 + 4 + . . . . . 97 + 98 + 99 + 100) = (-1)(K + J)
= -K - J

[Reveal] Spoiler:
C

Cheers,
Brent
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Brent Hanneson – Founder of gmatprepnow.com

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Joined: 14 Nov 2016
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Location: Malaysia
If J = 2 + 4 + 6 + 8 + . . . 98 + 100, and K = 1 + 3 + 5 + 7 + . . . + [#permalink]

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11 May 2017, 21:52
GMATPrepNow wrote:
If $$J = 2 + 4 + 6 + 8 + . . . 98 + 100$$, and $$K = 1 + 3 + 5 + 7 + . . . + 97 + 99$$, then $$1^2 - 2^2 + 3^2 - 4^2 + 5^2 - 6^2 + . . . . . + 97^2 - 98^2 + 99^2 - 100^2 =$$

A) J² - K²
B) -50(J² - K²)
C) -K - J
D) K² - J²
E) (-J - K)²

*kudos for all correct solutions

From Matt (Veritas Prep)

We could also cheat with a pattern:

$$n² - (n + 1)² => n² - (n² + 2n + 1) => -(2n + 1) => -(n + n + 1)$$ for any value of n.

Since we've got 1² - 2² + 3² - 4² ..., we've really got -(1 + 2) -(3 + 4) .... -(99 + 100), or -1 -2 -3 -4 .... - 99 - 100, or -(1 + 2 + 3 + ... + 100), or -(K + J), or -K - J.
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If J = 2 + 4 + 6 + 8 + . . . 98 + 100, and K = 1 + 3 + 5 + 7 + . . . + [#permalink]

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11 May 2017, 22:28
1
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GMATPrepNow wrote:
If J = 2 + 4 + 6 + 8 + . . . 98 + 100, and K = 1 + 3 + 5 + 7 + . . . + 97 + 99, then $$1^2 - 2^2 + 3^2 - 4^2 + 5^2 - 6^2 + . . . . . + 97^2 - 98^2 + 99^2 - 100^2 =$$

A) J² - K²
B) -50(J² - K²)
C) -K - J
D) K² - J²
E) (-J - K)²

*kudos for all correct solutions

There can be many solutions possible. But we will go with the basic solution, though it might be lengthy one to understand the concepts..
We can use tricks to solve problem as suggested by other members in exam.. Learning tricks is also very important.

J = 2+4+6+8+....+98+100
K = 1+3+5+7+....++97+99

$$1^2 - 2^2 + 3^2 - 4^2 + 5^2 - 6^2 + . . . . . + 97^2 - 98^2 + 99^2 - 100^2 =$$
= (1-2)(1+2) + (3-4) (3+4)+ (5-6)(5+6) +.......+ (97-98)(97+98) +(99-100)(99+100)
= -1[(1+2)+(3+4)(5+6) +......+ (97+98)+(99+100)]
= -1[ (1+3+5+...+97+99) + (2+4+6+...+98+100)]
= -1(K+J)
= -K-J

[Reveal] Spoiler:
Anwer : C

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Joined: 28 Mar 2013
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GMAT 1: 680 Q49 V32
Re: If J = 2 + 4 + 6 + 8 + . . . 98 + 100, and K = 1 + 3 + 5 + 7 + . . . + [#permalink]

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11 May 2017, 22:39
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We can pair any 2 consecutive term and apply (a+b) (a-b) in stead of a^2 - b ^2.

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Re: If J = 2 + 4 + 6 + 8 + . . . 98 + 100, and K = 1 + 3 + 5 + 7 + . . . +   [#permalink] 11 May 2017, 22:39
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