The sum of the series `(2)^2+2(4)^2+3(6)^2+....` upto 10 terms is

Options

A. 11300
B. 12100
C. 12300
D. 11200

Answer

To find the sum of the series \(2^2 + 2(4^2) + 3(6^2) + \ldots\) up to 10 terms, we can follow these steps:

### Step 1: Identify the General Term
The series can be expressed in terms of a general term. The pattern shows that the \(n\)-th term of the series can be written as:
\[ T_n = n \cdot (2n)^2 \]
This simplifies to:
\[ T_n = n \cdot 4n^2 = 4n^3 \]

### Step 2: Write the Sum of the First 10 Terms
We need to find the sum of the first 10 terms of the series:
\[ S_{10} = T_1 + T_2 + T_3 + \ldots + T_{10} = 4(1^3 + 2^3 + 3^3 + \ldots + 10^3) \]

### Step 3: Use the Formula for the Sum of Cubes
The formula for the sum of the first \(n\) cubes is:
\[ \sum_{k=1}^{n} k^3 = \left(\frac{n(n+1)}{2}\right)^2 \]
For \(n = 10\):
\[ \sum_{k=1}^{10} k^3 = \left(\frac{10 \cdot 11}{2}\right)^2 = (55)^2 = 3025 \]

### Step 4: Calculate the Total Sum
Now substituting this back into our expression for \(S_{10}\):
\[ S_{10} = 4 \cdot 3025 = 12100 \]

### Final Answer
Thus, the sum of the series up to 10 terms is:
\[ \boxed{12100} \]