Find the coefficient of `x^10` in the expansion of `(1+2x)^21 + (1 + 2x)^22 +……+ (1+ 2x)^30`

Answer

To find the coefficient of \( x^{10} \) in the expansion of \( (1 + 2x)^{21} + (1 + 2x)^{22} + \ldots + (1 + 2x)^{30} \), we can follow these steps:

### Step 1: Identify the general term in the binomial expansion
The general term in the expansion of \( (1 + 2x)^n \) is given by:
\[ T_k = \binom{n}{k} (2x)^k = \binom{n}{k} 2^k x^k \]
We need to find the coefficient of \( x^{10} \), which corresponds to \( k = 10 \).

### Step 2: Write the expression for the coefficient of \( x^{10} \)
For each term from \( n = 21 \) to \( n = 30 \), the coefficient of \( x^{10} \) is:
\[ \text{Coefficient of } x^{10} \text{ in } (1 + 2x)^n = \binom{n}{10} 2^{10} \]
Thus, for our expression, we have:
\[ \text{Coefficient of } x^{10} = \sum_{n=21}^{30} \binom{n}{10} 2^{10} \]

### Step 3: Factor out \( 2^{10} \)
Since \( 2^{10} \) is a common factor, we can factor it out:
\[ \text{Coefficient of } x^{10} = 2^{10} \sum_{n=21}^{30} \binom{n}{10} \]

### Step 4: Simplify the summation using binomial coefficient properties
We can use the identity:
\[ \sum_{k=r}^{m} \binom{k}{r} = \binom{m+1}{r+1} \]
In our case, we need to adjust the indices:
\[ \sum_{n=21}^{30} \binom{n}{10} = \sum_{k=10}^{30} \binom{k}{10} - \sum_{k=10}^{20} \binom{k}{10} \]
Using the identity:
\[ \sum_{k=10}^{30} \binom{k}{10} = \binom{31}{11} \]
and
\[ \sum_{k=10}^{20} \binom{k}{10} = \binom{21}{11} \]
Thus, we have:
\[ \sum_{n=21}^{30} \binom{n}{10} = \binom{31}{11} - \binom{21}{11} \]

### Step 5: Substitute back into the expression
Now substituting back into our expression for the coefficient:
\[ \text{Coefficient of } x^{10} = 2^{10} \left( \binom{31}{11} - \binom{21}{11} \right) \]

### Final Result
Thus, the coefficient of \( x^{10} \) in the expansion of \( (1 + 2x)^{21} + (1 + 2x)^{22} + \ldots + (1 + 2x)^{30} \) is:
\[ 2^{10} \left( \binom{31}{11} - \binom{21}{11} \right) \]