If `sum_(r=1)^(n)r(r+1)+sum_(r=1)^(n)(r+1)(r+2)=(n(an^(2)+bn+c))/(3),(a,b,cinN)` then (A) `a + b + c = 24` (B) `2a + b - c = 0` (C) `a - b + c= 6` (D) `ab + bc + ca = 11`

Options

A. `a + b + c = 24`
B. `2a + b - c = 0`
C. `a - b + c= 6`
D. `ab + bc + ca = 11`

Explanation

To solve the given problem, we need to analyze the left-hand side of the equation and simplify it step by step.

### Step 1: Write down the sums
We start with the expression:
\[ \sum_{r=1}^{n} r(r+1) + \sum_{r=1}^{n} (r+1)(r+2) \]

### Step 2: Simplify the first sum
The first sum can be simplified as follows:
\[ \sum_{r=1}^{n} r(r+1) = \sum_{r=1}^{n} (r^2 + r) = \sum_{r=1}^{n} r^2 + \sum_{r=1}^{n} r \]
Using the formulas for the sums:
- \(\sum_{r=1}^{n} r = \frac{n(n+1)}{2}\)
- \(\sum_{r=1}^{n} r^2 = \frac{n(n+1)(2n+1)}{6}\)

Thus,
\[ \sum_{r=1}^{n} r(r+1) = \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2} \]
To combine these, we need a common denominator:
\[ = \frac{n(n+1)(2n+1)}{6} + \frac{3n(n+1)}{6} = \frac{n(n+1)(2n+1 + 3)}{6} = \frac{n(n+1)(2n+4)}{6} \]

### Step 3: Simplify the second sum
Now, simplify the second sum:
\[ \sum_{r=1}^{n} (r+1)(r+2) = \sum_{r=1}^{n} (r^2 + 3r + 2) = \sum_{r=1}^{n} r^2 + 3\sum_{r=1}^{n} r + \sum_{r=1}^{n} 2 \]
Using the same formulas:
\[ = \frac{n(n+1)(2n+1)}{6} + 3 \cdot \frac{n(n+1)}{2} + 2n \]
Combining these:
\[ = \frac{n(n+1)(2n+1)}{6} + \frac{3n(n+1)}{2} + 2n \]
To combine these, we again need a common denominator:
\[ = \frac{n(n+1)(2n+1)}{6} + \frac{9n(n+1)}{6} + \frac{12n}{6} = \frac{n(n+1)(2n+1 + 9) + 12n}{6} \]
This simplifies to:
\[ = \frac{n(n+1)(2n + 10)}{6} \]

### Step 4: Combine both sums
Now we combine both results:
\[ \sum_{r=1}^{n} r(r+1) + \sum_{r=1}^{n} (r+1)(r+2) = \frac{n(n+1)(2n + 4)}{6} + \frac{n(n+1)(2n + 10)}{6} \]
Factoring out \(\frac{n(n+1)}{6}\):
\[ = \frac{n(n+1)}{6} \left( (2n + 4) + (2n + 10) \right) = \frac{n(n+1)}{6} (4n + 14) = \frac{n(n+1)(2n + 7)}{3} \]

### Step 5: Set equal to the right-hand side
According to the problem statement:
\[ \frac{n(n+1)(2n + 7)}{3} = \frac{n(an^2 + bn + c)}{3} \]
This implies:
\[ n(n+1)(2n + 7) = n(an^2 + bn + c) \]
Thus, we can equate coefficients:
- Coefficient of \(n^2\): \(a = 2\)
- Coefficient of \(n\): \(b = 9\)
- Constant term: \(c = 13\)

### Step 6: Verify the options
Now we check the options given:
1. \(a + b + c = 2 + 9 + 13 = 24\) (True)
2. \(2a + b - c = 2(2) + 9 - 13 = 4 + 9 - 13 = 0\) (True)
3. \(a - b + c = 2 - 9 + 13 = 6\) (True)
4. \(ab + bc + ca = 2 \cdot 9 + 9 \cdot 13 + 2 \cdot 13 = 18 + 117 + 26 = 161\) (False)

Thus, options A, B, and C are correct.