Let `T_(n) = sum_(r=1)^(n) (n)/(r^(2)-2r.n+2n^(2)), S_(n) = sum_(r=0)^(n)(n)/(r^(2)-2r.n+2n^(2))` then

Options

A. `T_(n) gt S_(n) AA n in N`
B. `T_(n) gt pi/4`
C. `S_(n) lt pi/4`
D. `underset(nrarroo)("lim")S_(n) = pi/4`

Answer

To solve the problem, we need to analyze the sums \( T_n \) and \( S_n \) defined as follows:

\[ T_n = \sum_{r=1}^{n} \frac{n}{r^2 - 2rn + 2n^2} \]
\[ S_n = \sum_{r=0}^{n} \frac{n}{r^2 - 2rn + 2n^2} \]

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

First, we rewrite \( T_n \):

\[ T_n = \sum_{r=1}^{n} \frac{n}{r^2 - 2rn + 2n^2} \]

Notice that the denominator can be factored:

\[ r^2 - 2rn + 2n^2 = (r - n)^2 + n^2 \]

Thus, we can rewrite \( T_n \):

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

### Step 2: Analyzing \( S_n \)

Now, let's analyze \( S_n \):

\[ S_n = \sum_{r=0}^{n} \frac{n}{r^2 - 2rn + 2n^2} \]

We can separate the first term where \( r = 0 \):

\[ S_n = \frac{n}{0^2 - 2 \cdot 0 \cdot n + 2n^2} + \sum_{r=1}^{n} \frac{n}{r^2 - 2rn + 2n^2} \]

Calculating the first term:

\[ \frac{n}{2n^2} = \frac{1}{2n} \]

Thus, we have:

\[ S_n = \frac{1}{2n} + T_n \]

### Step 3: Comparing \( T_n \) and \( S_n \)

Now we have:

\[ S_n = \frac{1}{2n} + T_n \]

To compare \( T_n \) and \( S_n \):

\[ T_n = S_n - \frac{1}{2n} \]

From this, we can see that:

\[ T_n = S_n - \frac{1}{2n} < S_n \]

since \( \frac{1}{2n} > 0 \).

### Conclusion

Thus, we conclude that:

\[ T_n < S_n \]

### Final Answer

The correct relationship is \( T_n < S_n \).