sabrina used a calcutator and started adding the whole numbers in order 1+2+3+4+5+..... what is the last number she would add that would get the sum on her calculator over 1000?
Try it yourself. You'll be surprised that it won't take very long to add up to 1,000.
To find the sum of the numbers from 1 to "n":
(1+n)(n/2)=sum
1001=(n^2+1)/2
2002=n^2+1
2001=n^2
n=44.7.....
So, 45 is the last number she would add.
995
To find the last number Sabrina would add to get a sum over 1000, we can use the formula for the sum of an arithmetic series.
The formula to find the sum of the numbers from 1 to "n" is:
Sum = (n/2)(first term + last term)
In this case, the first term is 1 and we want the sum to be over 1000.
So we have:
Sum > 1000
(n/2)(1 + last term) > 1000
Simplifying the equation, we get:
(n/2)(last term) > 999
Since we are only adding whole numbers, Sabrina can only add whole numbers as well. So, let's find the largest whole number for which the equation is true.
We can start with the number 1 and keep adding the next whole numbers until the equation is true.
For example:
1 + 2 = 3
1 + 2 + 3 = 6
1 + 2 + 3 + 4 = 10
1 + 2 + 3 + 4 + 5 = 15
and so on...
Eventually, we will find that the sum exceeds 1000. The last number Sabrina would add is the one that makes the sum just exceed 1000.
Using the formula for the sum of an arithmetic series, we can solve for "n".
(1 + n)(n/2) = 1000
n^2 + n = 2000
n^2 + n - 2000 = 0
Solving the quadratic equation, we find that n ≈ 44.7
Since n should be a whole number, the largest whole number that is less than or equal to 44.7 is 45.
Therefore, the last number Sabrina would add to get a sum over 1000 is 45.
To find the last number Sabrina would add, we need to calculate the sum of the whole numbers in order until the sum exceeds 1000.
One way to do this is by manually adding each number until we exceed 1000. However, this can be time-consuming and prone to errors.
Another method is to use a formula to calculate the sum of a series of whole numbers. The formula for the sum of the numbers from 1 to "n" is:
sum = (1 + n) * (n / 2)
We can set up an equation using this formula. We want the sum to be greater than 1000, so we solve for "n":
(1 + n) * (n / 2) > 1000
Multiplying both sides by 2 to remove the fraction:
(1 + n) * n > 2000
Expanding the expression:
n + n^2 > 2000
Rearranging the equation:
n^2 + n - 2000 > 0
To solve this quadratic inequality, we can find the roots of the equation:
n^2 + n - 2000 = 0
Using the quadratic formula, we get:
n = (-1 ± sqrt(1 + 4 * 2000)) / 2
Simplifying:
n = (-1 ± sqrt(8001)) / 2
Since we're looking for a positive value for "n", we take the positive square root:
n = (-1 + sqrt(8001)) / 2
Evaluating this expression:
n ≈ 44.7
Since "n" represents the last whole number Sabrina would add, we round up to the nearest whole number:
n ≈ 45
Therefore, the last number Sabrina would add to exceed a sum of 1000 is 45.