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tricky VIAC (Variables in the answer choices) question

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by Brent@GMATPrepNow » Mon May 08, 2017 6:07 am
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

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by Brent@GMATPrepNow » Wed May 10, 2017 1:05 pm
Brent@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
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: C

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Brent
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by Matt@VeritasPrep » Thu May 11, 2017 8:34 pm
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
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by Matt@VeritasPrep » Thu May 11, 2017 8:37 pm
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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by hazelnut01 » Thu May 11, 2017 9:44 pm
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
Hi Matt, How do we know J = K + 50?
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