If J = 2 + 4 + 6 + 8 + . . . 98 + 100, and K = 1 + 3 + 5 + 7 + . . . + 97 + 99, then 1² - 2² + 3² - 4² + 5² - 6² + . . . . . + 97² - 98² + 99² - 100² =
A) J² - K²
B) -50(J² - K²)
C) -K - J
D) K² - J²
E) (-J - K)²
Answer: C
Source: www.gmatprepnow.com
Difficulty level: 700
tricky VIAC (Variables in the answer choices) question
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We have several differences of squares hiding in the expression 1² - 2² + 3² - 4² + 5² - 6² + . . . . . + 97² - 98² + 99² - 100²Brent@GMATPrepNow wrote:If J = 2 + 4 + 6 + 8 + . . . 98 + 100, and K = 1 + 3 + 5 + 7 + . . . + 97 + 99, then 1² - 2² + 3² - 4² + 5² - 6² + . . . . . + 97² - 98² + 99² - 100² =
A) J² - K²
B) -50(J² - K²)
C) -K - J
D) K² - J²
E) (-J - K)²
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: C
Cheers,
Brent
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I'm thinking ...
1² - 2² + 3² - 4² ± ... + 99² - 100² =>
(1² + 3² + 5² + ... + 99²) - (2² + 4² + ... + 100²) =>
(1² + 3² + 5² + ... + 99²) - ((1+1)² + (3+1)² + ... + (99+1)²) =>
(1² + 3² + 5² + ... + 99²) - (1² + 3² + ... + 99² + 2*1 + 2*3 + ... + 2*99 + 1 + 1 + ... + 1) =>
(1² + 3² + 5² + ... + 99²) - ((1² + 3² + 5² + ... + 99²) + 2*(1 + 3 + ... + 99) + 50) =>
-2*(1 + 3 + ... + 99) - 50 =>
-2*K - 50 =>
-K - K - 50
And since J = K + 50, we can write this as
-K - (K + 50)
or
-K - J
1² - 2² + 3² - 4² ± ... + 99² - 100² =>
(1² + 3² + 5² + ... + 99²) - (2² + 4² + ... + 100²) =>
(1² + 3² + 5² + ... + 99²) - ((1+1)² + (3+1)² + ... + (99+1)²) =>
(1² + 3² + 5² + ... + 99²) - (1² + 3² + ... + 99² + 2*1 + 2*3 + ... + 2*99 + 1 + 1 + ... + 1) =>
(1² + 3² + 5² + ... + 99²) - ((1² + 3² + 5² + ... + 99²) + 2*(1 + 3 + ... + 99) + 50) =>
-2*(1 + 3 + ... + 99) - 50 =>
-2*K - 50 =>
-K - K - 50
And since J = K + 50, we can write this as
-K - (K + 50)
or
-K - J
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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.
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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Hi Matt, How do we know J = K + 50?Matt@VeritasPrep wrote:I'm thinking ...
1² - 2² + 3² - 4² ± ... + 99² - 100² =>
(1² + 3² + 5² + ... + 99²) - (2² + 4² + ... + 100²) =>
(1² + 3² + 5² + ... + 99²) - ((1+1)² + (3+1)² + ... + (99+1)²) =>
(1² + 3² + 5² + ... + 99²) - (1² + 3² + ... + 99² + 2*1 + 2*3 + ... + 2*99 + 1 + 1 + ... + 1) =>
(1² + 3² + 5² + ... + 99²) - ((1² + 3² + 5² + ... + 99²) + 2*(1 + 3 + ... + 99) + 50) =>
-2*(1 + 3 + ... + 99) - 50 =>
-2*K - 50 =>
-K - K - 50
And since J = K + 50, we can write this as
-K - (K + 50)
or
-K - J