1-2-3-4+...+1997-1998+1999-2000=
A) -2000
B) -1000
C) -1
D) 0
I think you have a typo and meant
1 - 2 + 3 - 4 + ... + 1997 - 1998 + 1999 - 2000
(all odds are positive and all evens are negative)
take them in pairs:
1-2 = -1
3-4 = -1
5-6 = -1
..
1999-2000=-1
so you have 1000 of these pairs, and the sum is -1000
To find the sum of the given sequence, let's group the terms with the same sign together:
(1 - 2) + (3 - 4) + ... + (1997 - 1998) + 1999 - 2000
The terms inside the parentheses will give us pairs with a sum of -1:
-1 - 1 - 1 - ... - 1
Since there are 999 pairs, the sum of the terms inside the parentheses is -999.
Now let's add the last two terms:
-999 + 1999 - 2000
Combining like terms:
(-999 + 1999) - 2000
1000 - 2000
-1000
Therefore, the sum of the given sequence is -1000. So the correct option is B) -1000.
To find the sum of the given series, you can use the concept of arithmetic progression.
First, let's look at the pattern of the series. It alternates between addition and subtraction, and each term increases by 1. We can see that the first term is 1, the second term is 2, the third term is 3, and so on.
To find the sum, we need to determine the number of terms in the series and then apply the formula for the sum of an arithmetic progression.
The formula for the sum of an arithmetic progression is:
S = (n/2)(a + l)
Where:
S = sum of the arithmetic progression
n = number of terms in the series
a = first term
l = last term
Now, we need to find the number of terms in the series. We can see that the series starts from 1 and ends at 1999, with a common difference of 1. Thus, the number of terms (n) can be calculated using the formula:
n = (l - a)/d + 1
Where:
d = common difference
In this case, the first term (a) is 1, the last term (l) is 1999, and the common difference (d) is 1. Plugging these values into the formula, we get:
n = (1999 - 1)/1 + 1
n = 1999
Now we have the number of terms (n). Let's apply the formula for the sum of an arithmetic progression:
S = (n/2)(a + l)
S = (1999/2)(1 + 1999)
S = 1999/2 * 2000
S = 999 * 2000
S = 1,998,000
So, the sum of the given series 1-2-3-4+...+1997-1998+1999-2000 is 1,998,000.
None of the answer choices A), B), C), or D) match the calculated sum.