If `sum_(r=1)^(n) r(r+1) = ((n+a)(n+b)(n+c))/(3)`, where `a gt b gt c`, then

Options

A. `2b = c`
B. `a^(3)-8b^(3)+c^(3) = 8abc`
C. a is prime number
D. `(a-2b)^(2) = 0`

Answer

To solve the problem, we need to evaluate the summation \( \sum_{r=1}^{n} r(r+1) \) and express it in the form given in the question.

### Step 1: Rewrite the Summation
The term \( r(r+1) \) can be rewritten as:
\[ r(r+1) = r^2 + r \]
Thus, we can express the summation as:
\[ \sum_{r=1}^{n} r(r+1) = \sum_{r=1}^{n} (r^2 + r) = \sum_{r=1}^{n} r^2 + \sum_{r=1}^{n} r \]

### Step 2: Use Known Formulas for Summations
We know the formulas for the summation of the first \( n \) natural numbers and the summation of the squares of the first \( n \) natural numbers:
- The sum of the first \( n \) natural numbers:
\[ \sum_{r=1}^{n} r = \frac{n(n+1)}{2} \]
- The sum of the squares of the first \( n \) natural numbers:
\[ \sum_{r=1}^{n} r^2 = \frac{n(n+1)(2n+1)}{6} \]

### Step 3: Substitute the Formulas into the Summation
Substituting these formulas into our expression, we get:
\[ \sum_{r=1}^{n} r(r+1) = \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2} \]

### Step 4: Simplify the Expression
To combine the two terms, we need a common denominator. The common denominator here is 6:
\[ \frac{n(n+1)(2n+1)}{6} + \frac{3n(n+1)}{6} = \frac{n(n+1)(2n+1 + 3)}{6} = \frac{n(n+1)(2n + 4)}{6} \]
Now, we can factor out a 2 from \( 2n + 4 \):
\[ = \frac{n(n+1) \cdot 2(n + 2)}{6} = \frac{n(n+1)(n+2)}{3} \]

### Step 5: Compare with the Given Expression
The problem states that:
\[ \sum_{r=1}^{n} r(r+1) = \frac{(n+a)(n+b)(n+c)}{3} \]
From our derived expression, we have:
\[ \frac{n(n+1)(n+2)}{3} \]
This implies:
\[ (n+a)(n+b)(n+c) = n(n+1)(n+2) \]

### Step 6: Expand the Right Side
Expanding \( (n+a)(n+b)(n+c) \):
\[ = n^3 + (a+b+c)n^2 + (ab+ac+bc)n + abc \]
And expanding \( n(n+1)(n+2) \):
\[ = n^3 + 3n^2 + 2n \]

### Step 7: Set Coefficients Equal
Now we can equate coefficients:
1. \( a + b + c = 3 \)
2. \( ab + ac + bc = 2 \)
3. \( abc = 0 \)

### Step 8: Solve for \( a, b, c \)
Since \( abc = 0 \), one of \( a, b, c \) must be zero. Let’s assume \( c = 0 \):
- Then \( a + b = 3 \)
- \( ab = 2 \)

The numbers \( a \) and \( b \) can be solved using these equations. The roots of the quadratic \( x^2 - 3x + 2 = 0 \) are \( x = 1 \) and \( x = 2 \). Thus, we can assign:
- \( a = 2 \)
- \( b = 1 \)
- \( c = 0 \)

### Conclusion
Thus, we have \( a = 2 \), \( b = 1 \), and \( c = 0 \) with \( a > b > c \).