To simplify the given summation expression:

$\sum_{n=m^2+1}^{(m+1)^2} k$

we follow these steps:

1. Identify the limits of the summation.
2. Recognize the summation formula for the series of integers.

Step-by-Step Solution:

1. Identify the limits of the summation:

   • Lower limit: $n = m^2 + 1$
   • Upper limit: $n = (m+1)^2$

2. Range of the summation:

   • The sequence of $k$ starts from $k = m^2 + 1$ and ends at $k = (m+1)^2$.

3. Number of terms in the summation: $(m+1)^2 - (m^2 + 1) + 1 = (m^2 + 2m + 1) - m^2 - 1 + 1 = 2m + 1$

4. Sum of integers from $m^2 + 1$ to $(m+1)^2$:

   • The sum of an arithmetic series $k$ where $k$ starts from $a$ and ends at $b$ is given by: $\text{Sum} = \frac{\text{number of terms}}{2} \times (\text{first term} + \text{last term})$
   • In our case, $a = m^2 + 1$ and $b = (m+1)^2$.

5. Calculate the sum: $\text{Sum} = \frac{2m+1}{2} \times (m^2 + 1 + (m+1)^2)$ Simplify $(m+1)^2$: $(m+1)^2 = m^2 + 2m + 1$ So the sum becomes: $\text{Sum} = \frac{2m+1}{2} \times (m^2 + 1 + m^2 + 2m + 1) = \frac{2m+1}{2} \times (2m^2 + 2m + 2)$ Simplify the expression inside the parenthesis: $2m^2 + 2m + 2 = 2(m^2 + m + 1)$ Therefore: $\text{Sum} = \frac{2m+1}{2} \times 2(m^2 + m + 1) = (2m+1)(m^2 + m + 1)$

So the simplified expression is:

$\sum_{n=m^2+1}^{(m+1)^2} k = (2m+1)(m^2 + m + 1)$

Would you like any further details or have any questions? Here are some related questions you might find interesting:

1. How to derive the sum of an arithmetic series?
2. What are common summation formulas?
3. How to simplify summations involving polynomial terms?
4. How to handle summations with variable limits?
5. What are the applications of summation in calculus?

Tip: Always verify the range and number of terms in a summation to ensure accurate results.