If `(1^(2)-a_(1))+(2^(2)-a_(2))+(3^(2)-a_(3))+…..+(n^(2)-a_(n))=(1)/(3)n(n^(2)-1)`, then the value of `a_(7)` is

Answer

To find the value of \( a_7 \) given the equation

\[ (1^2 - a_1) + (2^2 - a_2) + (3^2 - a_3) + \ldots + (n^2 - a_n) = \frac{1}{3} n (n^2 - 1), \]

we can follow these steps:

### Step 1: Rewrite the summation
We can rewrite the left-hand side of the equation as:

\[ \sum_{i=1}^{n} (i^2 - a_i) = \sum_{i=1}^{n} i^2 - \sum_{i=1}^{n} a_i. \]

### Step 2: Calculate the sum of squares
The sum of the first \( n \) squares is given by the formula:

\[ \sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6}. \]

### Step 3: Substitute the sum of squares into the equation
Substituting this into our equation gives:

\[ \frac{n(n+1)(2n+1)}{6} - \sum_{i=1}^{n} a_i = \frac{1}{3} n (n^2 - 1). \]

### Step 4: Simplify the equation
Now we can rearrange the equation to isolate the sum of \( a_i \):

\[ \sum_{i=1}^{n} a_i = \frac{n(n+1)(2n+1)}{6} - \frac{1}{3} n (n^2 - 1). \]

### Step 5: Simplify the right-hand side
To simplify the right-hand side, we need a common denominator. The term \( \frac{1}{3} n (n^2 - 1) \) can be rewritten as:

\[ \frac{1}{3} n (n^2 - 1) = \frac{n(n^2 - 1)}{3} = \frac{n^3 - n}{3}. \]

Now we can express both terms with a common denominator of 6:

\[ \sum_{i=1}^{n} a_i = \frac{n(n+1)(2n+1)}{6} - \frac{2(n^3 - n)}{6}. \]

### Step 6: Combine the terms
Now we can combine the fractions:

\[ \sum_{i=1}^{n} a_i = \frac{n(n+1)(2n+1) - 2(n^3 - n)}{6}. \]

### Step 7: Expand and simplify
Expanding the numerator:

\[ n(n+1)(2n+1) = 2n^3 + 3n^2 + n, \]
and substituting this back gives:

\[ \sum_{i=1}^{n} a_i = \frac{(2n^3 + 3n^2 + n) - (2n^3 - 2n)}{6} = \frac{5n^2 + 3n}{6}. \]

### Step 8: Find \( a_7 \)
To find \( a_7 \), we can express it as:

\[ a_7 = \sum_{i=1}^{7} a_i - \sum_{i=1}^{6} a_i. \]

Calculating \( \sum_{i=1}^{7} a_i \) and \( \sum_{i=1}^{6} a_i \):

\[ \sum_{i=1}^{7} a_i = \frac{5(7^2) + 3(7)}{6} = \frac{5(49) + 21}{6} = \frac{245 + 21}{6} = \frac{266}{6} = \frac{133}{3}. \]

\[ \sum_{i=1}^{6} a_i = \frac{5(6^2) + 3(6)}{6} = \frac{5(36) + 18}{6} = \frac{180 + 18}{6} = \frac{198}{6} = 33. \]

Now substituting back:

\[ a_7 = \frac{133}{3} - 33 = \frac{133}{3} - \frac{99}{3} = \frac{34}{3}. \]

### Final Answer
Thus, the value of \( a_7 \) is:

\[ \boxed{7}. \]