`3.6+6.9+9.12+....+3n(3n+3)=`

Options

A. `b=n,c=n`
B. b=n+1, c=n+2
C. b=n, c=n+1
D. b=n+1, c=n+1

Answer

To solve the problem \(3.6 + 6.9 + 9.12 + \ldots + 3n(3n + 3)\), we first need to identify the general term of the series.

### Step 1: Identify the general term
The series can be expressed in terms of \(n\):
- The first term is \(3 \cdot 2 = 6\)
- The second term is \(6 \cdot 3 = 18\)
- The third term is \(9 \cdot 4 = 36\)

From this, we can see that the \(r\)-th term can be written as:
\[ T_r = 3r(3r + 3) \]
This simplifies to:
\[ T_r = 9r^2 + 9r \]

### Step 2: Write the sum of the series
We need to find the sum of the first \(n\) terms:
\[ S_n = \sum_{r=1}^{n} T_r = \sum_{r=1}^{n} (9r^2 + 9r) \]
This can be separated into two sums:
\[ S_n = 9 \sum_{r=1}^{n} r^2 + 9 \sum_{r=1}^{n} r \]

### Step 3: Use the formulas for the sums
We use the formulas for the sums of the first \(n\) natural numbers and the sum of the squares of the first \(n\) natural numbers:
\[ \sum_{r=1}^{n} r = \frac{n(n + 1)}{2} \]
\[ \sum_{r=1}^{n} r^2 = \frac{n(n + 1)(2n + 1)}{6} \]

### Step 4: Substitute the formulas into the sum
Substituting these formulas into our expression for \(S_n\):
\[ S_n = 9 \left(\frac{n(n + 1)(2n + 1)}{6}\right) + 9 \left(\frac{n(n + 1)}{2}\right) \]

### Step 5: Simplify the expression
Now, we can simplify:
\[ S_n = \frac{9n(n + 1)(2n + 1)}{6} + \frac{9n(n + 1)}{2} \]
To combine these fractions, we need a common denominator, which is 6:
\[ S_n = \frac{9n(n + 1)(2n + 1)}{6} + \frac{27n(n + 1)}{6} \]
Now, factor out \(\frac{9n(n + 1)}{6}\):
\[ S_n = \frac{9n(n + 1)}{6} \left(2n + 1 + 3\right) \]
This simplifies to:
\[ S_n = \frac{9n(n + 1)}{6} (2n + 4) \]
Factoring out the 2 from \(2n + 4\):
\[ S_n = \frac{9n(n + 1)}{6} \cdot 2(n + 2) \]
This gives us:
\[ S_n = \frac{3n(n + 1)(n + 2)}{1} \]

### Final Result
Thus, the sum of the series is:
\[ S_n = 3n(n + 1)(n + 2) \]