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umg
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What is the value of 1^2 - 2^2 + 3^2 - 4^2 19 May 2017, 05:27
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E
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What is the value of

\(1^2 - 2^2 + 3^2 - 4^2 +\) ... \(+ 97^2 - 98^2 + 99^2 - 100^2\)

A. 5050
B. 5000
C. 0
D. -5000
E. -5050

Show SpoilerA neat Trick
Expand using formula \(a^2 - b^2\)
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E
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What is the value of 1^2 - 2^2 + 3^2 - 4^2 19 May 2017, 06:16
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What is the value of

\(1^2 - 2^2 + 3^2 - 4^2 +\) ... \(+ 97^2 - 98^2 + 99^2 - 100^2\)

A. 5050
B. 5000
C. 0
D. -5000
E. -5050
Show more

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)

To find the sum 1 + 2 + 3 + 4 + . . . . . 97 + 98 + 99 + 100, we can apply a nice formula:
1 + 2 + 3 + . . . n = (n)(n+1)/2

So, in this case, n = 100
So, 1 + 2 + 3 + 4 + . . . . . 97 + 98 + 99 + 100 = (100)(101)/2 = 5050

So, (-1)(1 + 2 + 3 + 4 + . . . . . 97 + 98 + 99 + 100) = (-1)(5050)
= -5050
Answer:
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E

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What is the value of 1^2 - 2^2 + 3^2 - 4^2 30 Oct 2020, 01:39
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1^2 - 2^2 = 1 - 4 = -3
3^2 - 4^2 = 9 - 16 = -7
5^2 - 6^2 = 25 - 36 = -11
.
.
99^2 - 100^2 = (99 - 100)(99 + 100) = -1 * 199 = -199 [a^2 - b^2 = (a + b)(a - b)]

The new series of differences formed is an A.P. with a common difference of -4: -3, -7, -11, -15, .......-199.

=> -199 = -3 + (n - 1) * (-4)
=> n = 50

Sum: \(\frac{n }{2}\) * [first term + last term]

=> Sum: \(\frac{50 }{2}\) * [-3 -199]

=> Sum: -5050

Answer E
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What is the value of 1^2 - 2^2 + 3^2 - 4^2 05 Nov 2020, 12:36
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What is the value of

\(1^2 - 2^2 + 3^2 - 4^2 +\) ... \(+ 97^2 - 98^2 + 99^2 - 100^2\)

A. 5050
B. 5000
C. 0
D. -5000
E. -5050

Show SpoilerA neat Trick
Expand using formula \(a^2 - b^2\)
Show more

Solution:

Let’s re-express the original expression as:

(1^2 - 2^2) + (3^2 - 4^2) + (5^2 - 6^2) + … + (97^2 - 98^2) + (99^2 - 100^2)

Note that each expression in parentheses is a difference of squares, so we can re-express as:

(1 - 2)(1 + 2) + (3 - 4)(3 + 4) + (5 - 6)(5 + 6) + … + (97 - 98)(97 + 98) + (99 -100)(99 + 100)

Now notice that the first factor of each term is -1:

(-1)(1 + 2) + (-1)(3 + 4) + (-1)(5 + 6) + … + (-1)(97 + 98) + (-1)(99 + 100)

Factoring out -1 from each term, we have:

(-1)(1 + 2 + 3 + 4 + 5 + 6 + … + 97 + 98 + 99 + 100)

We see that the second factor is the sum of the first 100 positive integers. We can evaluate this sum by using the formula: sum = average x number. The average of this evenly-spaced set is (1 + 100) / 2 = 101/2 = 50.5. We see that there are 100 numbers in the set, so we calculate that the sum of the first 100 positive integers is:

sum = average x number

sum = 50.5 x 100

sum = 5050

We recall that we factored out -1 prior to determining the sum, and we must use it to calculate the final answer. Thus, we have (-1) x 5050 = -5050.

Answer: E
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What is the value of 1^2 - 2^2 + 3^2 - 4^2 16 Dec 2025, 09:43
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\(1^2 - 2^2+3^2-4^2+...+99^2-100^2\)
\(= (1^2-100^2)-(2^2-99^2)+...-(50^2-51^2)\)
\(= (1+100)(1-100)-(2+99)(2-99)+...-(50+51)(50-51)\)
\( = (101)(-99)-(101)(-97)+...-(101)(-1) \)
\(= 101[-99+97-95+93+...-3+1]\)
\(= 101[(-2)+(-2)+...+(-2)]\) there are twenty-five (-2)s.
\(= 101(-2)(25)\)
\(=-5050.\)
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