Let `S_(n)=sum_(r=1)^(n)((r^(4)_r^(3)n+r^(2)n^(2)+2n^(4))/(n^(5)))` and `T_(n)=sum_(r=0)^(n-1)((r^(4)+r^(3)n^(2)+2n^(4))/(n^(5))),n=1,2,3,`……….then

Options

A. `T_(n)gt167/60`
B. `T_(n)lt167/60`
C. `S_(n)gt167/60`
D. `S_(n)lt167/60`

Explanation

To solve the problem, we need to evaluate the sums \( S_n \) and \( T_n \) given by:

\[ S_n = \sum_{r=1}^{n} \frac{r^4 + r^3 n + 2n^4}{n^5} \]

\[ T_n = \sum_{r=0}^{n-1} \frac{r^4 + r^3 n^2 + 2n^4}{n^5} \]

Let's break down the steps to find \( S_n \) and \( T_n \).

### Step 1: Simplifying \( S_n \)

We can rewrite \( S_n \) as follows:

\[ S_n = \frac{1}{n^5} \sum_{r=1}^{n} (r^4 + r^3 n + 2n^4) \]

This can be separated into three sums:

\[ S_n = \frac{1}{n^5} \left( \sum_{r=1}^{n} r^4 + n \sum_{r=1}^{n} r^3 + 2n^4 \sum_{r=1}^{n} 1 \right) \]

### Step 2: Evaluating the individual sums

Using the formulas for the sums:

- \( \sum_{r=1}^{n} r^4 = \frac{n(n+1)(2n+1)(3n^2 + 3n - 1)}{30} \)
- \( \sum_{r=1}^{n} r^3 = \left( \frac{n(n+1)}{2} \right)^2 \)
- \( \sum_{r=1}^{n} 1 = n \)

We can substitute these into our expression for \( S_n \).

### Step 3: Substitute the sums into \( S_n \)

Substituting the sums into \( S_n \):

\[ S_n = \frac{1}{n^5} \left( \frac{n(n+1)(2n+1)(3n^2 + 3n - 1)}{30} + n \left( \frac{n(n+1)}{2} \right)^2 + 2n^4 n \right) \]

### Step 4: Simplifying \( T_n \)

Now, let's simplify \( T_n \):

\[ T_n = \frac{1}{n^5} \sum_{r=0}^{n-1} (r^4 + r^3 n^2 + 2n^4) \]

This can also be separated into three sums:

\[ T_n = \frac{1}{n^5} \left( \sum_{r=0}^{n-1} r^4 + n^2 \sum_{r=0}^{n-1} r^3 + 2n^4 \sum_{r=0}^{n-1} 1 \right) \]

### Step 5: Evaluating the individual sums for \( T_n \)

Using similar formulas for the sums:

- \( \sum_{r=0}^{n-1} r^4 = \frac{(n-1)n(2n-1)(3n^2 - 3n + 1)}{30} \)
- \( \sum_{r=0}^{n-1} r^3 = \left( \frac{(n-1)n}{2} \right)^2 \)
- \( \sum_{r=0}^{n-1} 1 = n \)

Substituting these sums into \( T_n \):

\[ T_n = \frac{1}{n^5} \left( \frac{(n-1)n(2n-1)(3n^2 - 3n + 1)}{30} + n^2 \left( \frac{(n-1)n}{2} \right)^2 + 2n^4 n \right) \]

### Step 6: Final Evaluation

Now, we can evaluate both \( S_n \) and \( T_n \) for specific values of \( n \) (like \( n=1, 2, 3 \)) to find their respective sums.