If `(1+2x +3x^2)^10 = a_0 +a_1x +a_2x^2 + ……+a_20x^20` then `a_1` = ?

Options

A. 10.5
B. 20
C. 10
D. 5.5

Answer

To find the coefficient \( a_1 \) in the expansion of \( (1 + 2x + 3x^2)^{10} \), we can use the Binomial Theorem. The Binomial Theorem states that:

\[ (a + b + c)^n = \sum_{i+j+k=n} \frac{n!}{i!j!k!} a^i b^j c^k \]

where \( a, b, c \) are terms in the expression, and \( i, j, k \) are non-negative integers that sum to \( n \).

### Step 1: Identify the terms
In our case, we have:
- \( a = 1 \)
- \( b = 2x \)
- \( c = 3x^2 \)
- \( n = 10 \)

### Step 2: Find the coefficient of \( x \)
To find \( a_1 \), we need the coefficient of \( x \) in the expansion. The only way to get \( x \) from the terms is by selecting \( 2x \) once and \( 1 \) the remaining times, since \( 3x^2 \) contributes \( x^2 \) which cannot contribute to \( x \).

### Step 3: Determine the number of ways to choose terms
We can choose \( 2x \) once and \( 1 \) nine times. The number of ways to choose one \( 2x \) from the 10 terms is given by the binomial coefficient:

\[ \binom{10}{1} = 10 \]

### Step 4: Calculate the contribution to \( a_1 \)
The contribution to \( a_1 \) from choosing \( 2x \) once is:

\[ \text{Coefficient of } x = 2 \times \binom{10}{1} = 2 \times 10 = 20 \]

### Conclusion
Thus, the value of \( a_1 \) is:

\[ \boxed{20} \]

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