Find the coefficient of `x^5` in the expansion of `1+(1+x)+ (1+x)^2 + … + (1+ x)^(10)`.

Answer

To find the coefficient of \(x^5\) in the expansion of \(1 + (1+x) + (1+x)^2 + \ldots + (1+x)^{10}\), we can follow these steps:

### Step 1: Identify the series
The given expression is a sum of the first 11 terms of the series \( (1+x)^n \) for \( n = 0 \) to \( n = 10 \). This can be expressed as:
\[ S = 1 + (1+x) + (1+x)^2 + \ldots + (1+x)^{10} \]

### Step 2: Use the formula for the sum of a geometric series
This series can be recognized as a geometric series where:
- The first term \( a = 1 \)
- The common ratio \( r = (1+x) \)
- The number of terms \( n = 11 \)

The sum of a geometric series can be calculated using the formula:
\[ S_n = \frac{a(r^n - 1)}{r - 1} \]
Substituting the values:
\[ S = \frac{1((1+x)^{11} - 1)}{(1+x) - 1} = \frac{(1+x)^{11} - 1}{x} \]

### Step 3: Expand the expression
Now we need to expand \( (1+x)^{11} \) using the Binomial Theorem, which states:
\[ (1+x)^n = \sum_{k=0}^{n} \binom{n}{k} x^k \]
For \( n = 11 \):
\[ (1+x)^{11} = \sum_{k=0}^{11} \binom{11}{k} x^k \]

### Step 4: Find the coefficient of \( x^5 \)
We need to find the coefficient of \( x^5 \) in the expression:
\[ S = \frac{(1+x)^{11} - 1}{x} \]
The term \( -1 \) does not contribute to \( x^5 \), so we focus on \( (1+x)^{11} \).

The coefficient of \( x^6 \) in \( (1+x)^{11} \) will give us the coefficient of \( x^5 \) in \( S \) because when we divide by \( x \), we reduce the power by 1.

Using the binomial coefficient:
\[ \text{Coefficient of } x^6 \text{ in } (1+x)^{11} = \binom{11}{6} \]

### Step 5: Calculate \( \binom{11}{6} \)
\[ \binom{11}{6} = \frac{11!}{6!(11-6)!} = \frac{11!}{6!5!} \]
Calculating this:
\[ = \frac{11 \times 10 \times 9 \times 8 \times 7}{5 \times 4 \times 3 \times 2 \times 1} = \frac{55440}{120} = 462 \]

### Conclusion
The coefficient of \( x^5 \) in the expansion of \( 1 + (1+x) + (1+x)^2 + \ldots + (1+x)^{10} \) is \( 462 \).

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