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If J = 2 + 4 + 6 + 8 + . . . 98 + 100, and K = 1 + 3 + 5 + 7 + . . . +
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08 May 2017, 06:06
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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 + . . . +
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10 May 2017, 13:07
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 Answer: Cheers, Brent
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If J = 2 + 4 + 6 + 8 + . . . 98 + 100, and K = 1 + 3 + 5 + 7 + . . . +
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08 May 2017, 06:53
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 = 14+916= 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




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Re: If J = 2 + 4 + 6 + 8 + . . . 98 + 100, and K = 1 + 3 + 5 + 7 + . . . +
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08 May 2017, 07:25
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 \(1^22^2\), \(3^24^2\), and so on till \(99^2100^2\).. 1^22^2=(12)(1+2)=1(1+2)=12... 3^24^2=(34)(3+4)=1(3+4)=34.. So the equation becomes 1234......99100=(1+2+3+4+...+99+100)= [(2+4+6....+98+100)+(1+3+5+...+97+99)]=[(j)+(k)]=jk C
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Re: If J = 2 + 4 + 6 + 8 + . . . 98 + 100, and K = 1 + 3 + 5 + 7 + . . . +
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08 May 2017, 09:52
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 We can break the problem into a^2  b ^2 = 1^2  2^2 = (1+2) (12) = 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: 3 711.....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 + . . . +
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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 + . . . +
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11 May 2017, 22:28
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 \(1^2  2^2 + 3^2  4^2 + 5^2  6^2 + . . . . . + 97^2  98^2 + 99^2  100^2 =\) = (12)(1+2) + (34) (3+4)+ (56)(5+6) +.......+ (9798)(97+98) +(99100)(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) = KJ
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Re: If J = 2 + 4 + 6 + 8 + . . . 98 + 100, and K = 1 + 3 + 5 + 7 + . . . +
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11 May 2017, 22:39
Answer is C . We can pair any 2 consecutive term and apply (a+b) (ab) 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 + . . . +
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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 Answer: Cheers, Brent Hi Brent, I did't understand the step 3. Can you please elaborate, how did you write (12)(1+2) = (1)(1 + 2) (34)(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 + . . . +
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01 May 2018, 07:16
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 Answer: Cheers, Brent Hi Brent, I did't understand the step 3. Can you please elaborate, how did you write (12)(1+2) = (1)(1 + 2) (34)(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 + . . . +
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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 (12)(1+2) = (1)(1 + 2) (34)(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.




Re: If J = 2 + 4 + 6 + 8 + . . . 98 + 100, and K = 1 + 3 + 5 + 7 + . . . + &nbs
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