The expansion `1+x,1+x+x^(2),1+x+x^(2)+x^(3),….,1+x+x^(2)+…+x^(20)` are multipled together and the terms of the product thus obtained are arranged in increasing powers of `x` in the form of `a_(0)+a_(1)x+a_(2)x^(2)+…` then,
The value of `(a_(r ))/(a_(n-r))`, where `n` is the degree of the product.

Options

A. (a) `2`
B. (b) `1`
C. (c) `1//2`
D. (d) depends on `r`

Explanation

To solve the problem, we need to analyze the given expansions and their product. The expansions are:

1. \(1 + x\)
2. \(1 + x + x^2\)
3. \(1 + x + x^2 + x^3\)
4. ...
5. \(1 + x + x^2 + \ldots + x^{20}\)

We will denote the product of these expansions as \(f(x)\).

### Step 1: Determine the product \(f(x)\)

The product can be expressed as:

\[ f(x) = (1 + x)(1 + x + x^2)(1 + x + x^2 + x^3) \cdots (1 + x + x^2 + \ldots + x^{20}) \]

### Step 2: Find the degree of the product

The degree of the product \(f(x)\) is determined by the highest power of \(x\) that can be formed when multiplying these expansions. The highest power in each expansion is:

- From \(1 + x\): highest power is \(1\)
- From \(1 + x + x^2\): highest power is \(2\)
- From \(1 + x + x^2 + x^3\): highest power is \(3\)
- ...
- From \(1 + x + x^2 + \ldots + x^{20}\): highest power is \(20\)

The total highest power of \(x\) in the product is:

\[ 1 + 2 + 3 + \ldots + 20 = \frac{20 \cdot (20 + 1)}{2} = \frac{20 \cdot 21}{2} = 210 \]

Thus, the degree \(n\) of the product \(f(x)\) is \(210\).

### Step 3: Coefficients in the expansion

The expansion can be expressed as:

\[ f(x) = a_0 + a_1 x + a_2 x^2 + \ldots + a_{210} x^{210} \]

### Step 4: Relationship between coefficients

From the properties of the polynomial, we can derive that:

\[ a_r = a_{210 - r} \]

This means that the coefficients are symmetric around the middle term.

### Step 5: Find the value of \(\frac{a_r}{a_{n-r}}\)

Given that \(n = 210\), we have:

\[ \frac{a_r}{a_{n - r}} = \frac{a_r}{a_{210 - r}} = 1 \]

### Final Answer

Thus, the value of \(\frac{a_r}{a_{n - r}}\) is:

\[ \boxed{1} \]