December 20 , 2018
December 20 , 2018
10:00 PM PST
11:00 PM PST
This is the most inexpensive and attractive price in the market. Get the course now!
December 22 , 2018
December 22 , 2018
07:00 AM PST
09:00 AM PST
Attend this webinar to learn how to leverage Meaning and Logic to solve the most challenging Sentence Correction Questions.
Author
Message
TAGS:
CEO
Joined: 11 Sep 2015
Posts: 3243
Location: Canada
If J = 2 + 4 + 6 + 8 + . . . 98 + 100, and K = 1 + 3 + 5 + 7 + . . . +
[#permalink ]
Show Tags
D
E
Add Notes
Question Stats:
41% (02:29) correct
59% (02:21) wrong
based on 247 sessions
Hide Show timer Statistics
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
Official Answer and Stats are available only to registered users.
Register /
Login .
_________________
Test confidently with gmatprepnow.com
CEO
Joined: 11 Sep 2015
Posts: 3243
Location: Canada
Re: If J = 2 + 4 + 6 + 8 + . . . 98 + 100, and K = 1 + 3 + 5 + 7 + . . . +
[#permalink ]
Show Tags
GMATPrepNow wrote:
If J = 2 + 4 + 6 + 8 + . . . 98 + 100, and K = 1 + 3 + 5 + 7 + . . . + 97 + 99, then \(1^2 - 2^2 + 3^2 - 4^2 + 5^2 - 6^2 + . . . . . + 97^2 - 98^2 + 99^2 - 100^2 =\)
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
Answer:
Cheers,
Brent
_________________
Test confidently with gmatprepnow.com
VP
Joined: 05 Mar 2015
Posts: 1003
If J = 2 + 4 + 6 + 8 + . . . 98 + 100, and K = 1 + 3 + 5 + 7 + . . . +
[#permalink ]
Show Tags
GMATPrepNow wrote:
If J = 2 + 4 + 6 + 8 + . . . 98 + 100, and K = 1 + 3 + 5 + 7 + . . . + 97 + 99, then \(1^2 - 2^2 + 3^2 - 4^2 + 5^2 - 6^2 + . . . . . + 97^2 - 98^2 + 99^2 - 100^2 =\)
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: 7112
Re: If J = 2 + 4 + 6 + 8 + . . . 98 + 100, and K = 1 + 3 + 5 + 7 + . . . +
[#permalink ]
Show Tags
GMATPrepNow wrote:
If J = 2 + 4 + 6 + 8 + . . . 98 + 100, and K = 1 + 3 + 5 + 7 + . . . + 97 + 99, then \(1^2 - 2^2 + 3^2 - 4^2 + 5^2 - 6^2 + . . . . . + 97^2 - 98^2 + 99^2 - 100^2 =\)
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
_________________
1) 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 effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html
Director
Joined: 14 Nov 2014
Posts: 634
Location: India
GPA: 3.76
Re: If J = 2 + 4 + 6 + 8 + . . . 98 + 100, and K = 1 + 3 + 5 + 7 + . . . +
[#permalink ]
Show Tags
GMATPrepNow wrote:
If J = 2 + 4 + 6 + 8 + . . . 98 + 100, and K = 1 + 3 + 5 + 7 + . . . + 97 + 99, then \(1^2 - 2^2 + 3^2 - 4^2 + 5^2 - 6^2 + . . . . . + 97^2 - 98^2 + 99^2 - 100^2 =\)
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
Senior SC Moderator
Joined: 14 Nov 2016
Posts: 1324
Location: Malaysia
If J = 2 + 4 + 6 + 8 + . . . 98 + 100, and K = 1 + 3 + 5 + 7 + . . . +
[#permalink ]
Show Tags
GMATPrepNow wrote:
If \(J = 2 + 4 + 6 + 8 + . . . 98 + 100\), and \(K = 1 + 3 + 5 + 7 + . . . + 97 + 99\), then \(1^2 - 2^2 + 3^2 - 4^2 + 5^2 - 6^2 + . . . . . + 97^2 - 98^2 + 99^2 - 100^2 =\)
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.
_________________
"Be challenged at EVERY MOMENT." “Strength doesn’t come from what you can do. It comes from overcoming the things you once thought you couldn’t.” "Each stage of the journey is crucial to attaining new heights of knowledge." Rules for posting in verbal forum Advanced Search : https://gmatclub.com/forum/advanced-search/
Director
Joined: 13 Mar 2017
Posts: 678
Location: India
Concentration: General Management, Entrepreneurship
GPA: 3.8
WE: Engineering (Energy and Utilities)
If J = 2 + 4 + 6 + 8 + . . . 98 + 100, and K = 1 + 3 + 5 + 7 + . . . +
[#permalink ]
Show Tags
GMATPrepNow wrote:
If J = 2 + 4 + 6 + 8 + . . . 98 + 100, and K = 1 + 3 + 5 + 7 + . . . + 97 + 99, then \(1^2 - 2^2 + 3^2 - 4^2 + 5^2 - 6^2 + . . . . . + 97^2 - 98^2 + 99^2 - 100^2 =\)
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
_________________
CAT 2017 99th percentiler : VA 97.27 | DI-LR 96.84 | QA 98.04 | OA 98.95
MBA Social Network : WebMaggu Appreciate by Clicking +1 Kudos ( Lets be more generous friends.)
What I believe is : "Nothing is Impossible, Even Impossible says I'm Possible" : "Stay Hungry, Stay Foolish".
Intern
Joined: 28 Mar 2013
Posts: 4
Re: If J = 2 + 4 + 6 + 8 + . . . 98 + 100, and K = 1 + 3 + 5 + 7 + . . . +
[#permalink ]
Show Tags
Answer is C .
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
Intern
Joined: 05 Mar 2018
Posts: 3
Re: If J = 2 + 4 + 6 + 8 + . . . 98 + 100, and K = 1 + 3 + 5 + 7 + . . . +
[#permalink ]
Show Tags
GMATPrepNow wrote:
GMATPrepNow wrote:
If J = 2 + 4 + 6 + 8 + . . . 98 + 100, and K = 1 + 3 + 5 + 7 + . . . + 97 + 99, then \(1^2 - 2^2 + 3^2 - 4^2 + 5^2 - 6^2 + . . . . . + 97^2 - 98^2 + 99^2 - 100^2 =\)
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
Answer:
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...?
CEO
Joined: 11 Sep 2015
Posts: 3243
Location: Canada
Re: If J = 2 + 4 + 6 + 8 + . . . 98 + 100, and K = 1 + 3 + 5 + 7 + . . . +
[#permalink ]
Show Tags
MayurAgrawal wrote:
GMATPrepNow wrote:
GMATPrepNow wrote:
If J = 2 + 4 + 6 + 8 + . . . 98 + 100, and K = 1 + 3 + 5 + 7 + . . . + 97 + 99, then \(1^2 - 2^2 + 3^2 - 4^2 + 5^2 - 6^2 + . . . . . + 97^2 - 98^2 + 99^2 - 100^2 =\)
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
Answer:
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
_________________
Test confidently with gmatprepnow.com
Intern
Joined: 05 Mar 2018
Posts: 3
Re: If J = 2 + 4 + 6 + 8 + . . . 98 + 100, and K = 1 + 3 + 5 + 7 + . . . +
[#permalink ]
Show Tags
Quote:
Quote:
Hi Brent, I did't understand the step 3. Can you please elaborate, how did you write
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.
Re: If J = 2 + 4 + 6 + 8 + . . . 98 + 100, and K = 1 + 3 + 5 + 7 + . . . + &nbs
[#permalink ]
01 May 2018, 17:00
close
41% (02:29) correct 
