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
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
_________________
Brent Hanneson - founder of Greenlight Test Prep