## Sunday, May 23, 2010

### GR9768.62: Algebraic Expansion

62.  The coefficient of $x^3$ in the expansion of $(1+x)^3(2+x^2)^{10}$ is
(A)  $2^{14}$
(B)  $31$
(C)  $3 \choose 3$ + $10\choose 1$
(D)  $3 \choose 3$ + $2{10\choose 1}$
(E)  $3\choose 3$$10\choose 1$$2^9$

Solution:
First, observe that $(1+x)^3 = 1 + 3x + 3x^2 + x^3$.
And $(2+x^2)^{10} = {10\choose0}2^{10}(x^2)^0 + {10\choose1}2^{9} (x^2)^1 + {10\choose2} 2^8 (x^2)^2 \ldots$
$= {10\choose0}2^{10} + {10\choose1}2^{9} x^2 + {10\choose2} 2^8 x^4 \ldots$
$= 2^{10} + 10 \cdot 2^9 x^2 + 45 \cdot 2^8 x^4 + \dots$
Multiplying the two, then collect the terms with $x^3$:
$x^3 \cdot 2^{10}$, $3x \cdot 10 \cdot 2^9 x^2$.
The coefficient is:
$2^{10} + 30 \cdot 2^9 = 2^{10}(1 + 15) = 2^{10} \cdot 16 = 2^{10} \cdot 2^4 = 2^{14}$.
-sg-

#### 1 comment:

1. This question made me feel retarded. I calculated the answer as 16*2^10, and then I quit because that wasn't one of the available answers...