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

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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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Question Stats:

38% (02:36) correct 62% (02:15) wrong based on 264 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

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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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Top Contributor
8
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

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

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08 May 2017, 06:53
7
1
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
##### General Discussion
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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, 07:25
7
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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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
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
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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, 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
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

Anwer : C
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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
1
We can pair any 2 consecutive term and apply (a+b) (a-b) in stead of a^2 - b ^2.

Sent from my Moto G (4) using GMAT Club Forum mobile app
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Re: If J = 2 + 4 + 6 + 8 + . . . 98 + 100, and K = 1 + 3 + 5 + 7 + . . . +  [#permalink]

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30 Apr 2018, 20:01
GMATPrepNow wrote:
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)
= [color=red](-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

Cheers,
Brent

Hi Brent, I did't understand the step 3. Can you please elaborate, how did you write
(1-2)(1+2) = (-1)(1 + 2)
(3-4)(3+4) =(-1)(3+4)... and so on...?
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Re: If J = 2 + 4 + 6 + 8 + . . . 98 + 100, and K = 1 + 3 + 5 + 7 + . . . +  [#permalink]

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01 May 2018, 07:16
2
Top Contributor
MayurAgrawal wrote:
GMATPrepNow wrote:
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

Cheers,
Brent

Hi Brent, I did't understand the step 3. Can you please elaborate, how did you write
(1-2)(1+2) = (-1)(1 + 2)
(3-4)(3+4) =(-1)(3+4)... and so on...?

Sure thing.
Let's take (1 - 2)(1 + 2) + (3 - 4)(3 + 4) + (5 - 6)(5 + 6) + . . . . . + (97 - 98)(97 + 98) + (99 - 100)(99 + 100) and break it into its individual parts:
(1 - 2)(1 + 2) = (-1)(1 + 2) because 1 - 2 = -1
(3 - 4)(3 + 4) = (-1)(3 + 4) because 3 - 4 = -1
(5 - 6)(5 + 6) = (-1)(5 + 6) because 5 - 6 = -1
.
.
.

(97 - 98)(97 + 98) = (-1)(97 + 98) because 97 - 98 = -1
(99 - 100)(99 + 100) = (-1)(99 + 100) because 99 - 100 = -1

So, we get: (-1)(1 + 2) + (-1)(3 + 4) + (-1)(5 + 6) + . . . . . + (-1)(97 + 98) + (-1)(99 + 100)
From here, we can factor out the -1 to get: (-1)[(1 + 2) + (3 + 4) + (5 + 6) + . . . . . + (97 + 98) + (99 + 100)]
Which is the same as: (-1)(1 + 2 + 3 + 4 + . . . . . 97 + 98 + 99 + 100)

Does that help?

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

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01 May 2018, 17:00
Quote:
Quote:
Hi Brent, I did't understand the step 3. Can you please elaborate, how did you write
(1-2)(1+2) = (-1)(1 + 2)
(3-4)(3+4) =(-1)(3+4)... and so on...?

Sure thing.
Let's take (1 - 2)(1 + 2) + (3 - 4)(3 + 4) + (5 - 6)(5 + 6) + . . . . . + (97 - 98)(97 + 98) + (99 - 100)(99 + 100) and break it into its individual parts:
(1 - 2)(1 + 2) = (-1)(1 + 2) because 1 - 2 = -1
(3 - 4)(3 + 4) = (-1)(3 + 4) because 3 - 4 = -1
(5 - 6)(5 + 6) = (-1)(5 + 6) because 5 - 6 = -1
.
.
.

(97 - 98)(97 + 98) = (-1)(97 + 98) because 97 - 98 = -1
(99 - 100)(99 + 100) = (-1)(99 + 100) because 99 - 100 = -1

So, we get: (-1)(1 + 2) + (-1)(3 + 4) + (-1)(5 + 6) + . . . . . + (-1)(97 + 98) + (-1)(99 + 100)
From here, we can factor out the -1 to get: (-1)[(1 + 2) + (3 + 4) + (5 + 6) + . . . . . + (97 + 98) + (99 + 100)]
Which is the same as: (-1)(1 + 2 + 3 + 4 + . . . . . 97 + 98 + 99 + 100)

Does that help?

Cheers,
Brent

Silly Me. I totally missed subtraction. Thank you very much for clearing doubt.
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Re: If J = 2 + 4 + 6 + 8 + . . . 98 + 100, and K = 1 + 3 + 5 + 7 + . . . +  [#permalink]

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24 Dec 2019, 06:37
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

Brent's solution in this thread above is amazing and is the way to go. I am sharing here the approach for working backwards from the answer choices.

1. Take a sample from the two series (highlighted above)
$$J = 2 + 4 + 6 + 8 = 20$$
$$K = 1 + 3 + 5 + 7 = 16$$

Now, $$1^2 - 2^2 + 3^2 - 4^2 + 5^2 - 6^2 + 7^2 - 8^2 = -36$$

2. Plug $$J = 20$$, $$K = 16$$ in the choices to check which one yields $$-36$$
As you will realize that only C i.e. $$-K - J$$ works
$$-K - J = -20-16 = -36$$

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

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24 Dec 2019, 06:59
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 --> CORRECT: $$1^2 - 2^2 + 3^2 - 4^2 + 5^2 - 6^2 + . . . . . + 97^2 - 98^2 + 99^2 - 100^2$$ = $$(1^2 - 2^2) + (3^2 - 4^2) + (5^2 - 6^2) + . . . . . + (97^2 - 98^2) + (99^2 - 100^2)$$ = (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(J+K)=-J-K
D) K² - J²
E) (-J - K)²
Re: If J = 2 + 4 + 6 + 8 + . . . 98 + 100, and K = 1 + 3 + 5 + 7 + . . . +   [#permalink] 24 Dec 2019, 06:59
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