Which expressions are equal to [tex]$\sum_{n=1}^{50} n(n+6)$[/tex]?
Check all that apply.
[tex]$\sum_{n=1}^{50} n+\sum_{n=1}^{50}(n+6)$[/tex]
[tex]$\sum_{n=1}^{50} n^2+\sum_{n=1}^{50} 6 n$[/tex]
[tex]$\sum_{n=1}^{50} n \times \sum_{n=1}^{50}(n+6)$[/tex]
[tex]$\sum_{n=1}^{50} n^2+6 \sum_{n=1}^{50} n$[/tex]

Expand the original expression: ∑ n = 1 50 ​ n ( n + 6 ) = ∑ n = 1 50 ​ ( n 2 + 6 n ) .
Split the summation: ∑ n = 1 50 ​ ( n 2 + 6 n ) = ∑ n = 1 50 ​ n 2 + ∑ n = 1 50 ​ 6 n .
Simplify further: ∑ n = 1 50 ​ n 2 + ∑ n = 1 50 ​ 6 n = ∑ n = 1 50 ​ n 2 + 6 ∑ n = 1 50 ​ n .
The expressions equal to the original are: ∑ n = 1 50 ​ n 2 + ∑ n = 1 50 ​ 6 n and ∑ n = 1 50 ​ n 2 + 6 ∑ n = 1 50 ​ n .

Explanation

Understanding the Problem We are given the expression ∑ n = 1 50 ​ n ( n + 6 ) and we need to determine which of the given expressions are equal to it.

Expanding the Original Expression Let's expand the original expression: ∑ n = 1 50 ​ n ( n + 6 ) = ∑ n = 1 50 ​ ( n 2 + 6 n )

Splitting the Sum Now, we use the property of summation to split the sum: ∑ n = 1 50 ​ ( n 2 + 6 n ) = ∑ n = 1 50 ​ n 2 + ∑ n = 1 50 ​ 6 n

Taking the Constant Out We can also take the constant out of the summation: ∑ n = 1 50 ​ n 2 + ∑ n = 1 50 ​ 6 n = ∑ n = 1 50 ​ n 2 + 6 ∑ n = 1 50 ​ n

Analyzing the Given Expressions Now, let's analyze each of the given expressions:

∑ n = 1 50 ​ n + ∑ n = 1 50 ​ ( n + 6 ) = ∑ n = 1 50 ​ ( 2 n + 6 ) = 2 ∑ n = 1 50 ​ n + ∑ n = 1 50 ​ 6 = 2 ∑ n = 1 50 ​ n + 6 ( 50 ) = 2 ∑ n = 1 50 ​ n + 300 . This is not equal to the original expression.

∑ n = 1 50 ​ n 2 + ∑ n = 1 50 ​ 6 n = ∑ n = 1 50 ​ n 2 + 6 ∑ n = 1 50 ​ n . This is equal to the original expression.

∑ n = 1 50 ​ n × ∑ n = 1 50 ​ ( n + 6 ) = ∑ n = 1 50 ​ n × ( ∑ n = 1 50 ​ n + ∑ n = 1 50 ​ 6 ) = ∑ n = 1 50 ​ n × ( ∑ n = 1 50 ​ n + 300 ) . This is not equal to the original expression.

∑ n = 1 50 ​ n 2 + 6 ∑ n = 1 50 ​ n . This is equal to the original expression.

Final Answer Therefore, the expressions that are equal to ∑ n = 1 50 ​ n ( n + 6 ) are:


∑ n = 1 50 ​ n 2 + ∑ n = 1 50 ​ 6 n
∑ n = 1 50 ​ n 2 + 6 ∑ n = 1 50 ​ n
Examples
Understanding summations is crucial in many fields, such as physics and computer science. For example, when calculating the total energy of a system with multiple particles, you might need to sum the kinetic energy of each particle. Similarly, in computer science, when analyzing the complexity of an algorithm, you often need to sum the number of operations performed in each step. The ability to manipulate and simplify summations allows for efficient calculations and better understanding of complex systems.

Answered by GinnyAnswer | 2025-07-03

The expressions equal to ∑ n = 1 50 ​ n ( n + 6 ) are ∑ n = 1 50 ​ n 2 + ∑ n = 1 50 ​ 6 n and ∑ n = 1 50 ​ n 2 + 6 ∑ n = 1 50 ​ n . The other options do not match the original expression. This demonstrates how to expand and simplify summations effectively.
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Answered by Anonymous | 2025-07-04