Coefficient of `x^7` in `(1 + x)^10 + x(1 + x)^9 + x^2 (1 + x)^8+………..+x^10` is

Options

A. 210
B. 330
C. 420
D. 260

Explanation

To find the coefficient of \( x^7 \) in the expression

\[ (1 + x)^{10} + x(1 + x)^9 + x^2(1 + x)^8 + \ldots + x^{10}, \]

we can rewrite the expression in a more manageable form.

### Step 1: Rewrite the expression
The expression can be rewritten as:

\[ \sum_{k=0}^{10} x^k (1 + x)^{10-k}. \]

### Step 2: Expand the terms
We need to find the coefficient of \( x^7 \) in the entire sum. For each term \( x^k (1 + x)^{10-k} \), we can find the coefficient of \( x^{7-k} \) in \( (1 + x)^{10-k} \).

### Step 3: Use the binomial theorem
Using the binomial theorem, the coefficient of \( x^m \) in \( (1 + x)^n \) is given by \( \binom{n}{m} \). Therefore, the coefficient of \( x^{7-k} \) in \( (1 + x)^{10-k} \) is:

\[ \binom{10-k}{7-k} = \binom{10-k}{3}. \]

### Step 4: Sum the contributions
Now we need to sum these contributions for \( k = 0 \) to \( k = 7 \) (since \( 7-k \) must be non-negative):

\[ \sum_{k=0}^{7} \binom{10-k}{3}. \]

### Step 5: Change the index of summation
To simplify the summation, we can change the index of summation by letting \( j = 10 - k \). Then when \( k = 0 \), \( j = 10 \) and when \( k = 7 \), \( j = 3 \). Thus, we have:

\[ \sum_{j=3}^{10} \binom{j}{3}. \]

### Step 6: Use the hockey-stick identity
Using the hockey-stick identity in combinatorics, we have:

\[ \sum_{j=r}^{n} \binom{j}{r} = \binom{n+1}{r+1}. \]

In our case, \( r = 3 \) and \( n = 10 \):

\[ \sum_{j=3}^{10} \binom{j}{3} = \binom{11}{4}. \]

### Step 7: Calculate the binomial coefficient
Now we calculate \( \binom{11}{4} \):

\[ \binom{11}{4} = \frac{11!}{4!(11-4)!} = \frac{11 \times 10 \times 9 \times 8}{4 \times 3 \times 2 \times 1} = \frac{7920}{24} = 330. \]

### Conclusion
Thus, the coefficient of \( x^7 \) in the given expression is \( 330 \).