Math GRE - #10

What is the coefficient of $x^3$ in the expansion of $(1+x)^3(2+x^2)^{10}$?

• $2^{14}$
• $31$
• $\dbinom{3}{1}+\dbinom{10}{1}$
• $\dbinom{3}{1}+2\dbinom{10}{1}$
• $\dbinom{3}{3}\dbinom{10}{1}2^9$

Solution :

This is a straightforward application of the binomial theorem. We need to take into account of the coefficients of the constant term, $x$, $x^2$, and $x^3$ from both $(1+x)^3$ and $(2+x^2)^{10}$.
For $(1+x)^3$, we have:

• $c_1 = 1$
• $c_x=\dbinom{3}{1}=3$
• $c_{x^2}=\dbinom{3}{2}=3$
• $c_{x^3}=\dbinom{3}{3}=1$

For $(2+x)^10$, we have:

• $d_1=2^10$
• $d_x=0$
• $d_{x^2}=\dbinom{10}{1}2^9$
• $d_{x^3}=0$

Fortunately, $(2+x^2)^{10}$ does not have an $x$ or $x^3$ term.
Thus the coefficient of $x^3$ in the product is: \[\begin{eqnarray*}
c_1\cdot d_{x^3} + c_x\cdot d_{x^2} + c_{x^2} \cdot d_x + c_{x^3}\cdot d_1 & = & \dbinom{3}{1}\dbinom{10}{1}2^9 + \dbinom{3}{3}2^{10} \\
& = & 3\cdot 10 \cdot 2^9 + 2\cdot 2^9 \\
& = & 32\cdot 2^9 = 2^{14} \end{eqnarray*}\]