the question is :

Sn= sigma r=1 to n ,t(r) =1/6n(2n^2+9n+13).

Then find the value of sigma r=1 to n , [t(r)]^1/2

To tackle the problem, we first need to understand the expression given for \( S_n \), which is defined as the sum of \( t(r) \) from \( r = 1 \) to \( n \). The function \( t(r) \) is given by the formula \( t(r) = \frac{1}{6} n (2n^2 + 9n + 13) \). Our goal is to find the value of \( \sum_{r=1}^{n} [t(r)]^{1/2} \).

Breaking Down the Function

First, let's clarify what \( t(r) \) represents. The formula for \( t(r) \) is dependent on \( n \), not on \( r \). This means that for any fixed \( n \), \( t(r) \) will yield the same value for all \( r \) from 1 to \( n \). Therefore, we can simplify our calculations significantly.

Calculating \( t(r) \)

Substituting \( n \) into the function, we have:

• For any \( r \), \( t(r) = \frac{1}{6} n (2n^2 + 9n + 13) \).

This indicates that \( t(r) \) is constant for all \( r \) in the range from 1 to \( n \). Thus, we can denote this constant value as \( C \), where:

C = \frac{1}{6} n (2n^2 + 9n + 13)

Finding the Square Root

Next, we need to compute \( [t(r)]^{1/2} \). Since \( t(r) \) is constant, we can write:

[t(r)]^{1/2} = \sqrt{C} = \sqrt{\frac{1}{6} n (2n^2 + 9n + 13)}

Summing Up the Square Roots

Now, we can substitute this back into our summation:

\sum_{r=1}^{n} [t(r)]^{1/2} = \sum_{r=1}^{n} \sqrt{C} = n \cdot \sqrt{C}

This is because we are summing the same value \( \sqrt{C} \) for \( n \) terms.

Final Calculation

Now, substituting back the expression for \( C \), we have:

n \cdot \sqrt{\frac{1}{6} n (2n^2 + 9n + 13)}

Thus, the final result for the summation \( \sum_{r=1}^{n} [t(r)]^{1/2} \) is:

n \cdot \sqrt{\frac{1}{6} n (2n^2 + 9n + 13)}

In summary, the value of \( \sum_{r=1}^{n} [t(r)]^{1/2} \) simplifies nicely due to the constancy of \( t(r) \) across the range of \( r \). This approach not only streamlines the calculation but also highlights the importance of recognizing patterns in mathematical functions.