If `(1 ^(2) - t _(1)) + (2 ^(2) - t _(2)) + ......+ ( n ^(2) - t _(n)) =(1)/(3) n ( n ^(2) -1 ),` then `t _(n)` is

Answer

To solve the problem, we need to find the expression for \( t_n \) given the equation:

\[ (1^2 - t_1) + (2^2 - t_2) + \ldots + (n^2 - t_n) = \frac{1}{3} n (n^2 - 1) \]

### Step-by-Step Solution:

1. **Rewrite the Left Side**:
We can express the left side of the equation as:
\[ \sum_{r=1}^{n} (r^2 - t_r) = \sum_{r=1}^{n} r^2 - \sum_{r=1}^{n} t_r \]
Therefore, we can rewrite the equation as:
\[ \sum_{r=1}^{n} r^2 - \sum_{r=1}^{n} t_r = \frac{1}{3} n (n^2 - 1) \]

2. **Sum of Squares Formula**:
The formula for the sum of the squares of the first \( n \) natural numbers is:
\[ \sum_{r=1}^{n} r^2 = \frac{n(n + 1)(2n + 1)}{6} \]
Substituting this into our equation gives:
\[ \frac{n(n + 1)(2n + 1)}{6} - \sum_{r=1}^{n} t_r = \frac{1}{3} n (n^2 - 1) \]

3. **Simplifying the Right Side**:
The right side can be simplified as:
\[ \frac{1}{3} n (n^2 - 1) = \frac{1}{3} n (n - 1)(n + 1) \]

4. **Combine and Rearrange**:
Rearranging the equation gives:
\[ \sum_{r=1}^{n} t_r = \frac{n(n + 1)(2n + 1)}{6} - \frac{1}{3} n (n - 1)(n + 1) \]

5. **Finding a Common Denominator**:
The common denominator for the two fractions on the right side is 6. Thus, we can rewrite:
\[ \frac{1}{3} n (n - 1)(n + 1) = \frac{2n(n - 1)(n + 1)}{6} \]
Therefore, we have:
\[ \sum_{r=1}^{n} t_r = \frac{n(n + 1)(2n + 1)}{6} - \frac{2n(n - 1)(n + 1)}{6} \]

6. **Combine the Terms**:
Combine the fractions:
\[ \sum_{r=1}^{n} t_r = \frac{n(n + 1)}{6} \left( (2n + 1) - 2(n - 1) \right) \]
Simplifying the expression inside the parentheses:
\[ (2n + 1) - 2(n - 1) = 2n + 1 - 2n + 2 = 3 \]
Thus, we have:
\[ \sum_{r=1}^{n} t_r = \frac{n(n + 1) \cdot 3}{6} = \frac{n(n + 1)}{2} \]

7. **Finding \( t_n \)**:
To find \( t_n \), we can use the formula for the sum of the first \( n \) terms:
\[ t_n = \sum_{r=1}^{n} t_r - \sum_{r=1}^{n-1} t_r \]
We know:
\[ \sum_{r=1}^{n-1} t_r = \frac{(n-1)n}{2} \]
Therefore:
\[ t_n = \frac{n(n + 1)}{2} - \frac{(n-1)n}{2} = \frac{n(n + 1) - (n^2 - n)}{2} = \frac{n(n + 1 - n + 1)}{2} = \frac{n \cdot 2}{2} = n \]

### Final Result:
Thus, we conclude that:
\[ t_n = n \]