The coefficient of `x^(5)` in the expansion of
`1+(1+x)+(1+x)^(2)+……..+(1+x)^(10)` is equal to

Options

A. 462
B. 450
C. 543
D. 446

Answer

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

### Step 1: Rewrite the Series
The given series can be rewritten as:
\[ S = 1 + (1+x) + (1+x)^2 + \ldots + (1+x)^{10} \]
This is a geometric series where the first term \( a = 1 \) and the common ratio \( r = (1+x) \). The number of terms \( n = 11 \).

### Step 2: Use the Formula for the Sum of a Geometric Series
The sum of a geometric series can be calculated using the formula:
\[ S_n = \frac{a(1 - r^n)}{1 - r} \]
Applying this to our series:
\[ S = \frac{1 - (1+x)^{11}}{1 - (1+x)} = \frac{1 - (1+x)^{11}}{-x} = \frac{(1+x)^{11} - 1}{x} \]

### Step 3: Expand \( (1+x)^{11} \)
Using the Binomial Theorem, we can expand \( (1+x)^{11} \):
\[ (1+x)^{11} = \sum_{r=0}^{11} \binom{11}{r} x^r \]

### Step 4: Substitute into the Sum
Substituting this expansion back into our expression for \( S \):
\[ S = \frac{\sum_{r=0}^{11} \binom{11}{r} x^r - 1}{x} = \sum_{r=1}^{11} \binom{11}{r} x^{r-1} \]

### Step 5: Find the Coefficient of \( x^5 \)
To find the coefficient of \( x^5 \) in \( S \), we need to find the term where \( r-1 = 5 \), which means \( r = 6 \):
\[ \text{Coefficient of } x^5 = \binom{11}{6} \]

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

### Final Answer
Thus, the coefficient of \( x^5 \) in the expansion is \( 462 \).

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