Find the sum of the first 60 positive even integers.
The correct answer and explanation is:
Correct Answer:
The sum of the first 60 positive even integers is 3660.
Explanation:
To find the sum of the first 60 positive even integers, start by understanding what even numbers are. Even numbers are divisible by 2, so the first 60 positive even integers are:
2, 4, 6, 8, 10, …, up to the 60th term.
This forms an arithmetic sequence where:
- The first term a=2a = 2
- The common difference d=2d = 2
- The number of terms n=60n = 60
The formula for the sum SnS_n of the first nn terms in an arithmetic sequence is: Sn=n2⋅(2a+(n−1)d)S_n = \frac{n}{2} \cdot (2a + (n – 1)d)
Substitute the values into the formula: S60=602⋅[2⋅2+(60−1)⋅2]S_{60} = \frac{60}{2} \cdot [2 \cdot 2 + (60 – 1) \cdot 2] S60=30⋅[4+59⋅2]S_{60} = 30 \cdot [4 + 59 \cdot 2] S60=30⋅[4+118]S_{60} = 30 \cdot [4 + 118] S60=30⋅122=3660S_{60} = 30 \cdot 122 = 3660
Alternatively, use the simpler formula for the sum of the first nn even numbers: Sum=n(n+1)\text{Sum} = n(n + 1)
For n=60n = 60: Sum=60⋅61=3660\text{Sum} = 60 \cdot 61 = 3660
Both methods give the same correct result. This problem highlights the use of arithmetic sequences and offers a practical example of how patterns in numbers can be used to simplify calculations. Recognizing patterns in even numbers allows efficient problem solving and strengthens understanding of algebraic thinking and number properties. Arithmetic series formulas provide tools to quickly evaluate such sums without manual addition.