The sum first 19 terms of the series `1^2+2(2^2)+3^2+2(4^2)+5^2+...` is :

Options

A. 4200
B. 4410
C. 3800
D. 3610

Answer

To find the sum of the first 19 terms of the series \(1^2 + 2(2^2) + 3^2 + 2(4^2) + 5^2 + \ldots\), we can follow these steps:

### Step 1: Identify the pattern in the series
The series can be rewritten based on the pattern of the terms:
- The odd-indexed terms are \(1^2, 3^2, 5^2, \ldots\)
- The even-indexed terms are \(2(2^2), 2(4^2), \ldots\)

### Step 2: Rewrite the series
We can express the series as:
\[ 1^2 + 2(2^2) + 3^2 + 2(4^2) + 5^2 + \ldots \]
This can be rewritten as:
\[ 1^2 + 2^2 + 2^2 + 3^2 + 4^2 + 4^2 + 5^2 + \ldots \]
Where every even term \(2n\) contributes \(2n^2\) twice.

### Step 3: Separate the sums
We can separate the sums of odd and even indexed terms:
- Odd indexed terms: \(1^2, 3^2, 5^2, \ldots, 19^2\)
- Even indexed terms: \(2^2, 4^2, 6^2, \ldots, 18^2\)

### Step 4: Calculate the sum of odd indexed terms
The sum of squares of the first \(n\) odd numbers can be calculated using the formula:
\[ \text{Sum of squares of first } n \text{ odd numbers} = n^3 \]
For \(n = 10\) (since there are 10 odd numbers from 1 to 19):
\[ \text{Sum of odd indexed terms} = 10^3 = 1000 \]

### Step 5: Calculate the sum of even indexed terms
The even indexed terms can be expressed as:
\[ 2^2 + 4^2 + 6^2 + \ldots + 18^2 = 2^2(1^2 + 2^2 + 3^2 + \ldots + 9^2) \]
Using the formula for the sum of squares:
\[ \text{Sum of squares from } 1 \text{ to } n = \frac{n(n+1)(2n+1)}{6} \]
For \(n = 9\):
\[ \text{Sum} = \frac{9(10)(19)}{6} = 285 \]
Thus, the sum of even indexed terms becomes:
\[ 2^2 \times 285 = 4 \times 285 = 1140 \]

### Step 6: Combine the sums
Now, we add the sums of the odd and even indexed terms:
\[ \text{Total Sum} = 1000 + 1140 = 2140 \]

### Final Answer
The sum of the first 19 terms of the series is \(2140\).

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