`1+(1-1.2^2)+(1-3.4^2)+(1-5.6^2)+.....(1-19.20^2) = alpha-220beta` find`(alpha,beta)`

Options

A. `(10,103)`
B. `(11,103)`
C. `(10,107)`
D. `(11,97)`

Answer

To solve the problem \(1 + (1 - 1 \cdot 2^2) + (1 - 3 \cdot 4^2) + (1 - 5 \cdot 6^2) + \ldots + (1 - 19 \cdot 20^2) = \alpha - 220\beta\), we will break it down step by step.

### Step 1: Identify the series
The series can be expressed as:
\[ S = 1 + \sum_{n=1}^{10} \left(1 - (2n-1)(2n)^2\right) \]
This is because the odd numbers \(1, 3, 5, \ldots, 19\) correspond to \(2n-1\) where \(n\) ranges from \(1\) to \(10\) and the even numbers \(2, 4, 6, \ldots, 20\) correspond to \(2n\).

### Step 2: Simplify the series
We can separate the constant \(1\) from the summation:
\[ S = 1 + \sum_{n=1}^{10} \left(1 - (2n-1)(2n)^2\right) = 1 + \sum_{n=1}^{10} 1 - \sum_{n=1}^{10} (2n-1)(2n)^2 \]
The first summation gives \(10\) since there are \(10\) terms of \(1\):
\[ S = 1 + 10 - \sum_{n=1}^{10} (2n-1)(2n)^2 \]
Thus,
\[ S = 11 - \sum_{n=1}^{10} (2n-1)(2n)^2 \]

### Step 3: Calculate the summation
Now we need to calculate:
\[ \sum_{n=1}^{10} (2n-1)(2n)^2 = \sum_{n=1}^{10} (2n-1) \cdot 4n^2 = 4 \sum_{n=1}^{10} (2n^3 - n^2) \]
This can be split into two separate sums:
\[ = 4 \left(2\sum_{n=1}^{10} n^3 - \sum_{n=1}^{10} n^2\right) \]

### Step 4: Use formulas for summation
Using the formulas for the sums of cubes and squares:
\[ \sum_{n=1}^{k} n^3 = \left(\frac{k(k+1)}{2}\right)^2 \]
\[ \sum_{n=1}^{k} n^2 = \frac{k(k+1)(2k+1)}{6} \]
For \(k = 10\):
\[ \sum_{n=1}^{10} n^3 = \left(\frac{10 \cdot 11}{2}\right)^2 = 55^2 = 3025 \]
\[ \sum_{n=1}^{10} n^2 = \frac{10 \cdot 11 \cdot 21}{6} = 385 \]

### Step 5: Substitute back into the equation
Now substituting these values back:
\[ \sum_{n=1}^{10} (2n-1)(2n)^2 = 4 \left(2 \cdot 3025 - 385\right) = 4(6050 - 385) = 4 \cdot 5665 = 22660 \]

### Step 6: Final calculation of S
Now substituting back into \(S\):
\[ S = 11 - 22660 = -22649 \]

### Step 7: Compare with the given form
We have:
\[ -22649 = \alpha - 220\beta \]
This can be rearranged to find \(\alpha\) and \(\beta\).

### Step 8: Finding \(\alpha\) and \(\beta\)
To express \(-22649\) in the form \(\alpha - 220\beta\), we can set \(\alpha = -22649 + 220\beta\).

Choosing \(\beta = 103\):
\[ \alpha = -22649 + 220 \cdot 103 = -22649 + 22660 = 11 \]

Thus, we find:
\[ \alpha = 11, \quad \beta = 103 \]

### Final Answer
\((\alpha, \beta) = (11, 103)\)