Find the sum of the series :
`1*3^2 + 2* 5^2 + 3*7^2 + …` to 20 terms :

Options

A. a)12896
B. b)187898
C. c)98970
D. d)188090

Answer

To find the sum of the series \( S = 1 \cdot 3^2 + 2 \cdot 5^2 + 3 \cdot 7^2 + \ldots \) up to 20 terms, we first need to identify the general term of the series.

### Step 1: Identify the nth term
The series can be expressed in terms of \( n \):
- The coefficients are \( 1, 2, 3, \ldots, n \).
- The bases of the squares are \( 3, 5, 7, \ldots \), which can be represented as \( 2n + 1 \).

Thus, the nth term \( T_n \) of the series can be written as:
\[ T_n = n \cdot (2n + 1)^2 \]

### Step 2: Expand the nth term
Now, we expand \( T_n \):
\[ T_n = n \cdot (2n + 1)^2 = n \cdot (4n^2 + 4n + 1) = 4n^3 + 4n^2 + n \]

### Step 3: Write the sum of the first 20 terms
The sum \( S \) of the first 20 terms can be expressed as:
\[ S = \sum_{n=1}^{20} T_n = \sum_{n=1}^{20} (4n^3 + 4n^2 + n) \]
This can be separated into three sums:
\[ S = 4 \sum_{n=1}^{20} n^3 + 4 \sum_{n=1}^{20} n^2 + \sum_{n=1}^{20} n \]

### Step 4: Use formulas for summation
We will use the following formulas for summation:
1. \( \sum_{n=1}^{k} n = \frac{k(k + 1)}{2} \)
2. \( \sum_{n=1}^{k} n^2 = \frac{k(k + 1)(2k + 1)}{6} \)
3. \( \sum_{n=1}^{k} n^3 = \left( \frac{k(k + 1)}{2} \right)^2 \)

For \( k = 20 \):
- \( \sum_{n=1}^{20} n = \frac{20 \cdot 21}{2} = 210 \)
- \( \sum_{n=1}^{20} n^2 = \frac{20 \cdot 21 \cdot 41}{6} = 2870 \)
- \( \sum_{n=1}^{20} n^3 = \left( \frac{20 \cdot 21}{2} \right)^2 = 44100 \)

### Step 5: Substitute back into the sum
Now substituting these values back into the equation for \( S \):
\[ S = 4 \cdot 44100 + 4 \cdot 2870 + 210 \]
Calculating each term:
- \( 4 \cdot 44100 = 176400 \)
- \( 4 \cdot 2870 = 11480 \)

Now, adding these results:
\[ S = 176400 + 11480 + 210 = 188090 \]

### Final Answer
Thus, the sum of the series up to 20 terms is:
\[ \boxed{188090} \]