Question

The coefficient of $x^{12}$ in the expansion of $(x^2+2x+2)^8$ is

• A. 1120
• B. 2240
• C. 2576
• D. 4152

Hint:

For finding a specific coefficient in a multinomial expansion, write down the general term and the two main equations: one for the power of the variable and one for the sum of the exponents. Then solve the system of equations for all possible non-negative integer solutions.

Answer 1

The Correct Option is C

We need to find the coefficient of $x^{12}$ in the expansion of $(x^2+2x+2)^8$.
The general term in the multinomial expansion of $(a+b+c)^n$ is $\frac{n!}{p!q!r!} a^p b^q c^r$, where $p+q+r=n$.
Here, $a=x^2, b=2x, c=2$, and $n=8$. The general term is:
$\frac{8!}{p!q!r!} (x^2)^p (2x)^q (2)^r = \frac{8!}{p!q!r!} 2^{q+r} x^{2p+q}$.
We need the term with $x^{12}$, so we must have $2p+q=12$.
We also have the condition that $p, q, r$ are non-negative integers such that $p+q+r=8$.
From $2p+q=12$, we can see that $q$ must be an even number. Also, $q = 12-2p$. Substituting this into $p+q+r=8$: $p + (12-2p) + r = 8 \Rightarrow 12-p+r=8 \Rightarrow r = p-4$.
Since $r \geq 0$, we must have $p-4 \geq 0$, so $p \geq 4$. Since $q \geq 0$, we must have $12-2p \geq 0$, so $12 \geq 2p$, which means $p \leq 6$. Possible integer values for $p$ are 4, 5, 6.
Case 1: $p=4$. $q = 12 - 2(4) = 4$. $r = 4 - 4 = 0$. (Check: $p+q+r = 4+4+0=8$. Correct). Coefficient: $\frac{8!}{4!4!0!} 2^{4+0} = \frac{70}{1} \times 16 = 1120$.
Case 2: $p=5$. $q = 12 - 2(5) = 2$. $r = 5 - 4 = 1$. (Check: $p+q+r = 5+2+1=8$. Correct). Coefficient: $\frac{8!}{5!2!1!} 2^{2+1} = \frac{8 \times 7 \times 6}{2} \times 8 = 168 \times 8 = 1344$.
Case 3: $p=6$. $q = 12 - 2(6) = 0$. $r = 6 - 4 = 2$. (Check: $p+q+r = 6+0+2=8$. Correct). Coefficient: $\frac{8!}{6!0!2!} 2^{0+2} = \frac{8 \times 7}{2} \times 4 = 28 \times 4 = 112$.
The total coefficient of $x^{12}$ is the sum of the coefficients from these three cases.
Total coefficient = $1120 + 1344 + 112 = 2576$.