Find the sum of the series 1 + 2 + 4 + 5 + 7 + 8 + 10 + 11 + ... + 299, which is the sum of the integers from 0 to 300, omitting the multiples of 3.
well, I suspect that you can find the sum of the arithmetic sequence starting with 1 and ending at 299 with d = 1
then you need to find the sum of the arithmetic sequence
3 6 9 12 ....... 297
from 3 to 297 with d = 3
then subtract
see:
http://www.mathsisfun.com/algebra/sequences-sums-arithmetic.html
To find the sum of this series, we can break it down into three separate arithmetic series.
First, let's consider the series of numbers from 1 to 300. The sum of this series can be found using the formula for the sum of an arithmetic series:
S1 = (n/2)(first term + last term)
= (300/2)(1 + 300)
= 150(301)
= 45,150
Next, let's consider the series of multiples of 3 from 3 to 300. The number of terms in this series is given by:
n2 = (last term - first term)/common difference + 1
= (300 - 3)/3 + 1
= 98
The sum of this series can be found using the formula for the sum of an arithmetic series:
S2 = (n/2)(first term + last term)
= (98/2)(3 + 300)
= 49(303)
= 14,847
Finally, let's calculate the sum of the series of multiples of 3 from 0 to 300. We can find this by subtracting the sum of the series from 3 to 300 (S2) from the sum of the series from 1 to 300 (S1):
S3 = S1 - S2
= 45,150 - 14,847
= 30,303
Therefore, the sum of the given series 1 + 2 + 4 + 5 + 7 + 8 + 10 + 11 + ... + 299 is 30,303.
To find the sum of the series 1 + 2 + 4 + 5 + 7 + 8 + 10 + 11 + ... + 299, which omits the multiples of 3, we need to break down the series into two parts: the sum of all integers from 1 to 299 and the sum of all multiples of 3 from 3 to 300.
Step 1: Find the sum of all integers from 1 to 299
To find the sum of an arithmetic series, we can use the formula Sn = (n/2)(a + l), where Sn is the sum of the first n terms, a is the first term, and l is the last term.
In this case, the first term (a) is 1 and the last term (l) is 299. Therefore, the number of terms (n) is (299 - 1) + 1 = 299.
Using the formula Sn = (n/2)(a + l), we can calculate the sum of the series from 1 to 299:
S1-299 = (299/2)(1 + 299)
= (299/2)(300)
= 44850
So, the sum of all integers from 1 to 299 is 44850.
Step 2: Find the sum of all multiples of 3 from 3 to 300
To find the sum of an arithmetic series of multiples, we can use a similar formula. In this case, the first term (a) is 3, the last term (l) is 300, and the common difference (d) is 3.
The number of terms (n) can be found using the formula: n = (l - a)/d + 1
Thus, n = (300 - 3)/3 + 1 = 100.
Using the formula Sn = (n/2)(a + l), we can calculate the sum of the series from 3 to 300:
S3-300 = (100/2)(3 + 300)
= (100/2)(303)
= 15150
So, the sum of all multiples of 3 from 3 to 300 is 15150.
Step 3: Calculate the final sum
To get the sum of the series 1 + 2 + 4 + 5 + 7 + 8 + 10 + 11 + ... + 299, we subtract the sum of all multiples of 3 from the sum of all integers from 1 to 299:
Final Sum = Sum(1-299) - Sum(3-300) = 44850 - 15150 = 29700
Therefore, the sum of the given series is 29700.