The expression shown is:

$2 + 4 + 6 + \dots + 2n = n(n+1)$

This is a formula for the sum of the first $n$ even numbers. We can verify the formula by understanding that each term in the series is an even number, $2, 4, 6, \dots, 2n$, and we need to find the sum of these terms up to $2n$.

Explanation:

1. General Form of Sum of Even Numbers: The $k$-th even number can be written as $2k$.
2. Sum of Series: The series can be represented as: $2 + 4 + 6 + \dots + 2n = 2(1 + 2 + 3 + \dots + n)$
3. Sum of First $n$ Natural Numbers: The sum of the first $n$ natural numbers is: $1 + 2 + 3 + \dots + n = \frac{n(n+1)}{2}$
4. Final Calculation: Substituting this result back, we get: $2(1 + 2 + 3 + \dots + n) = 2 \cdot \frac{n(n+1)}{2} = n(n+1)$ which confirms the formula.

Thus, the sum of the first $n$ even numbers is indeed $n(n+1)$.

Would you like further details or have any questions?

Further Questions

1. What is the formula for the sum of the first $n$ odd numbers?
2. How can you generalize this approach to find the sum of multiples of a different integer, such as 3?
3. Can we use a similar approach to find the sum of squares of the first $n$ even numbers?
4. How does this formula relate to arithmetic series in general?
5. What other techniques can be used to verify this summation formula?

Tip

Remember that finding the sum of an arithmetic series often involves identifying a pattern and using known formulas for series, such as the sum of the first $n$ natural numbers.