In the expansion of `(x+x^(2)+x^(3)+…….) (1+x+x^(2)+x^(3)) (x^(2)+x^(3)+x^(4)+……+x^(10))` the coefficient of `x^(6)` is

Options

A. `10`
B. `18`
C. `30`
D. `12`

Explanation

To find the coefficient of \( x^6 \) in the expansion of

\[ (x + x^2 + x^3 + \ldots)(1 + x + x^2 + x^3)(x^2 + x^3 + x^4 + \ldots + x^{10}), \]

we will break down the problem step by step.

### Step 1: Simplify Each Series
1. The first series \( x + x^2 + x^3 + \ldots \) is a geometric series with first term \( x \) and common ratio \( x \). The sum can be expressed as:
\[ \frac{x}{1 - x} \quad \text{for } |x| < 1. \]

2. The second series \( 1 + x + x^2 + x^3 \) is also a geometric series with first term \( 1 \) and common ratio \( x \):
\[ \frac{1 - x^4}{1 - x} \quad \text{for } |x| < 1. \]

3. The third series \( x^2 + x^3 + x^4 + \ldots + x^{10} \) can be rewritten as:
\[ x^2(1 + x + x^2 + \ldots + x^8) = x^2 \cdot \frac{1 - x^9}{1 - x} \quad \text{for } |x| < 1. \]

### Step 2: Combine the Series
Now we can combine these series:
\[ \frac{x}{1 - x} \cdot \frac{1 - x^4}{1 - x} \cdot \left( x^2 \cdot \frac{1 - x^9}{1 - x} \right). \]

This simplifies to:
\[ \frac{x^3(1 - x^4)(1 - x^9)}{(1 - x)^3}. \]

### Step 3: Expand the Numerator
Next, we need to expand the numerator \( x^3(1 - x^4)(1 - x^9) \):
\[ x^3(1 - x^4 - x^9 + x^{13}) = x^3 - x^7 - x^{12} + x^{16}. \]

### Step 4: Find the Coefficient of \( x^6 \)
Now we need to find the coefficient of \( x^6 \) in the expression:
\[ \frac{x^3 - x^7 - x^{12} + x^{16}}{(1 - x)^3}. \]

The term \( x^3 \) in the numerator contributes to the coefficient of \( x^6 \) when combined with \( \frac{1}{(1 - x)^3} \).

### Step 5: Use the Binomial Theorem
The series \( \frac{1}{(1 - x)^3} \) can be expanded using the binomial theorem:
\[ \frac{1}{(1 - x)^3} = \sum_{n=0}^{\infty} \binom{n + 2}{2} x^n. \]

To find the coefficient of \( x^3 \) (since \( x^3 \) from the numerator needs \( x^3 \) from the denominator to make \( x^6 \)), we look for \( n = 3 \):
\[ \text{Coefficient of } x^3 = \binom{3 + 2}{2} = \binom{5}{2} = 10. \]

### Conclusion
Thus, the coefficient of \( x^6 \) in the expansion is \( 10 \).