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Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
If J = 2 + 4 + 6 + 8 + . . . 98 + 100, and K = 1 +
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10 May 2017, 13:11
10 May 2017, 13:11
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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
A) J² - K²
B) -50(J² - K²)
C) -K - J
D) K² - J²
E) (-J - K)²
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
Re: If J = 2 + 4 + 6 + 8 + . . . 98 + 100, and K = 1 +
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10 May 2017, 13:14
10 May 2017, 13:14
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GreenlightTestPrep 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
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:
Show: ::
C
Cheers,
Brent
General Discussion
Re: If J = 2 + 4 + 6 + 8 + . . . 98 + 100, and K = 1 +
[#permalink]
15 Feb 2023, 12:19
15 Feb 2023, 12:19
1
Hi Brent GreenlightTestPrep
= (1 - 2)(1 + 2) + (3 - 4)(3 + 4) +....
= (-1)[(1 + 2) + (3 + 4) + (5 + 6) + . . . . . + (97 + 98) + (99 + 100)]
If I multiply (-1) into (1 + 2) again, I'll get (-1-2) rather than (1 - 2)(1 + 2). Did I miss something here?
Could you clarify? Thanks Brent
= (1 - 2)(1 + 2) + (3 - 4)(3 + 4) +....
= (-1)[(1 + 2) + (3 + 4) + (5 + 6) + . . . . . + (97 + 98) + (99 + 100)]
If I multiply (-1) into (1 + 2) again, I'll get (-1-2) rather than (1 - 2)(1 + 2). Did I miss something here?
Could you clarify? Thanks Brent
