If the sum of the first 3𝑛 positive integers is 150 more than the sum of the first 𝑛 positive integers, then find the sum of the first 4𝑛 positive integers.
+23246 The formula for the sum of an arithmetic series is: Sum = n(a + l) / 2
where n = number of terms a = first term l = last term
The sum of the first 3n positive integers is: Sum = 3n(1 + 3n) / 2
The sum of the first n positive integers is: Sum = n(1 + n) / 2
The equation becomes: 3n(1 + 3n) / 2 = 150 + n(1 + n) / 2
Solving: 3n(1 + 3n) = 300 + n(1 + n)
3n + 9n2 = 300 + n + n2
8n2 + 2n - 300 = 0
4n2 + n - 150 = 0
(4n + 25)(n - 6) = 0
n = 6
The sum of the first 4n positive integers is: Sum = 4n(1 + 4n) / 2
Since 4n = 24, the sum is ...
