The problem asks us to express the sum of four consecutive even integers in terms of $n$, where $n$ is the first integer.

Consecutive even integers differ by 2. So, if $n$ is the first even integer, the next three even integers would be:

• First integer: $n$
• Second integer: $n + 2$
• Third integer: $n + 4$
• Fourth integer: $n + 6$

Now, to find the sum of these four integers, we simply add them together:

$n + (n + 2) + (n + 4) + (n + 6)$

Simplify the expression:

$n + n + 2 + n + 4 + n + 6 = 4n + 12$

So, the sum of the four consecutive even integers is $4n + 12$.

Final Answer:

The sum of four consecutive even integers in terms of $n$ is $4n + 12$.

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Would you like more details on any part of the explanation?

Here are 5 related questions to explore further:

1. What is the sum of four consecutive odd integers, starting with $n$?
2. How can you find the product of four consecutive even integers starting with $n$?
3. If the sum of four consecutive even integers is 100, what is the value of $n$?
4. How would the sum change if the integers were consecutive odd integers instead of even?
5. How can you generalize this approach for any number of consecutive even integers?

Tip: When dealing with sequences, it's helpful to define variables clearly and recognize consistent patterns in how they progress!