If `S_(n)=Sigma_(r=1)^(n)t_(r)=(1)/(6)n(2n^(2)+9n+13)`, then `Sigma_(r=1)^(n)sqrt(t_(r))` is equal to

Options

A. `(1)/(2)n(n+1)`
B. `(1)/(2)n(n+3)`
C. `(n+1)^(2)`
D. `n^(2)`

Answer

To solve the problem, we need to find the value of \( \Sigma_{r=1}^{n} \sqrt{t_r} \) given that \( S_n = \Sigma_{r=1}^{n} t_r = \frac{1}{6} n (2n^2 + 9n + 13) \).

### Step 1: Find \( t_n \)
We know that:
\[ S_n = \Sigma_{r=1}^{n} t_r \]
and
\[ S_{n-1} = \Sigma_{r=1}^{n-1} t_r \]
Thus, we have:
\[ t_n = S_n - S_{n-1} \]

### Step 2: Calculate \( S_{n-1} \)
Using the formula for \( S_n \):
\[ S_n = \frac{1}{6} n (2n^2 + 9n + 13) \]
we can find \( S_{n-1} \) by substituting \( n-1 \) into the formula:
\[ S_{n-1} = \frac{1}{6} (n-1) (2(n-1)^2 + 9(n-1) + 13) \]

### Step 3: Simplify \( S_{n-1} \)
Calculating \( S_{n-1} \):
\[ S_{n-1} = \frac{1}{6} (n-1) (2(n^2 - 2n + 1) + 9(n - 1) + 13) \]
\[ = \frac{1}{6} (n-1) (2n^2 - 4n + 2 + 9n - 9 + 13) \]
\[ = \frac{1}{6} (n-1) (2n^2 + 5n + 6) \]

### Step 4: Calculate \( t_n \)
Now, substituting \( S_n \) and \( S_{n-1} \) into the equation for \( t_n \):
\[ t_n = S_n - S_{n-1} \]
\[ = \frac{1}{6} n (2n^2 + 9n + 13) - \frac{1}{6} (n-1)(2n^2 + 5n + 6) \]

### Step 5: Simplify \( t_n \)
Expanding \( t_n \):
\[ t_n = \frac{1}{6} \left[ n(2n^2 + 9n + 13) - (n-1)(2n^2 + 5n + 6) \right] \]
Calculating the expression inside the brackets:
\[ = n(2n^2 + 9n + 13) - (2n^2 + 5n + 6)(n-1) \]
Expanding the second term:
\[ = n(2n^2 + 9n + 13) - (2n^3 + 5n^2 + 6n - 2n^2 - 5n - 6) \]
\[ = n(2n^2 + 9n + 13) - (2n^3 + 3n^2 + n + 6) \]

### Step 6: Collect like terms
Combining like terms will give us \( t_n \):
\[ t_n = \frac{1}{6} \left[ n(2n^2 + 9n + 13) - 2n^3 - 3n^2 - n - 6 \right] \]

### Step 7: Find \( \sqrt{t_r} \)
From our calculations, we find that:
\[ t_n = (n + 1)^2 \]
Thus,
\[ \sqrt{t_r} = r + 1 \]

### Step 8: Calculate \( \Sigma_{r=1}^{n} \sqrt{t_r} \)
Now we need to calculate:
\[ \Sigma_{r=1}^{n} \sqrt{t_r} = \Sigma_{r=1}^{n} (r + 1) \]
This can be separated into two sums:
\[ = \Sigma_{r=1}^{n} r + \Sigma_{r=1}^{n} 1 = \frac{n(n + 1)}{2} + n \]
\[ = \frac{n(n + 1)}{2} + \frac{2n}{2} = \frac{n(n + 1) + 2n}{2} = \frac{n^2 + 3n}{2} \]

### Final Result
Thus, the value of \( \Sigma_{r=1}^{n} \sqrt{t_r} \) is:
\[ \frac{n(n + 3)}{2} \]