1

Find the coefficient of $x^3$ in the expansion $(1+x)^3 \cdot (2+x^2)^{10}$.

I did the first part, which is expanding the second equation at $x^3$ and I got: $\binom {10} 3 \cdot 2^7 \cdot (x^2)^3 = 15360 (x^2)^3$, but I can't figure out what to do from here.

Alex M.
  • 35,207
1234567890
  • 11
  • Expand $(1+x)^3$ and $(2+x^2)^{10}$ but only for terms of degree $3$ or less in $x$, giving $(1+3x+3x^2+x^3)(2^{10}+2^9(45)x^2)$ and find the co-efficient of $x^3$ in this,but first observe that we can drop the terms "$1$" and "$3x^2$" from $1+3x+3x^2+x^3$ because they will contribute nothing to the co-efficient of $x^3$. – DanielWainfleet Dec 06 '15 at 21:05
  • ERROR In my first comment the term $2^9(45)x^2$ should be $2^9(10)x^2$. – DanielWainfleet Dec 06 '15 at 21:13

4 Answers4

4

Let's try it straight from the binomial theorem: $$ (1+x)^3\cdot(2+x^2)^{10} = \sum_{j=0}^3 {3 \choose j} x^{3-j} \cdot \sum_{k=0}^{10} {10 \choose k} 2^k (x^2)^{10-k}. $$ Good so far? Now, note these terms can be arranged as $$ \sum_{j=0}^3 \sum_{k=0}^{10} {3 \choose j} {10 \choose k}2^k x^{3-j} (x^2)^{10-k}. $$ Note the $x$ terms combine as $$ x^{23 - j - 2k}. $$ We want the exponent of $x$ to be $3$, which means we want $$ 23 - j - 2k = 3 \iff j + 2k = 20, $$ which is satisfied only when $j = 0$ and $k = 10$, and when $j = 2$ and $k = 9$. Thus the coefficient of $x^3$ is $$ {3 \choose 0}{10 \choose 10}2^{10} + {3 \choose 2}{10 \choose 9}2^{9} = 16384. $$

bcf
  • 3,120
0

Think of this as an integer composition: to get $x^3$, you can either get the coefficient of $x^3$ from the first term times the coefficient of $x^0$ or the constant from the second term, or find the coefficient of $x$ from the first term times the coefficient of $x^2$ from the second term. Any other combination will not give you $x^3$. From there, just use binomial theorem.

Kevin Long
  • 5,159
0

Here is a different way. Let $f(x)=(1+x)^3(2+x^2)^{10}$. Then \begin{align} f'(x)&=3(1+x)^2(2+x^2)^{10}+20x(1+x)^3(2+x^2)^9\\ f''(x)&=...\\ f'''(x)&=6 \left(x^2+2\right)^{10}+360 x (x+1) \left(x^2+2\right)^9+9 (x+1)^2 \left(20 \left(x^2+2\right)^9+360 x^2 \left(x^2+2\right)^8\right)+(x+1)^3 \left(1080 x \left(x^2+2\right)^8+5760 x^3 \left(x^2+2\right)^7\right) \end{align}

Hence the coefficient you are after is $$\frac16f'''(0)=16384$$

Math-fun
  • 9,507
0

It's also convenient to use the coefficient of operator $[x^n]$ to denote the coefficient of $x^n$ in an algebraic expression.

We obtain \begin{align*} [x^3](1+x)^3(2+x^2)^{10}&=[x^3]\left(\sum_{k=0}^{10}\binom{10}{k}x^{2k}2^{10-k}\right)(1+x)^3\\ &=[x^3]\left(\binom{10}{0}2^{10}x^0+\binom{10}{1}2^9x^2\right)(1+x)^3\tag{1}\\ &=\left(2^{10}[x^3]+10\cdot2^9[x^1]\right)\sum_{k=0}^{3}\binom{3}{x}x^k\tag{2}\\ &=\left(2^{10}\binom{3}{3}+10\cdot2^9\binom{3}{1}\right)\\ &=2^{10}+30\cdot 2^9\\ &=16384 \end{align*}

Comment:

  • In (1) we ignore all summands in the first sum with powers of $x$ greater than 3

  • In (2) we use the rule $[x^{k+l}]f(x)=[x^k]x^{-l}f(x), k,l\geq 0$ and the linearity of the coefficient of operator.

Markus Scheuer
  • 108,315