Find the coefficient of $x^6$ in $(2+2x+2x^2+2x^3+2x^4+x^5)^5$

Question

Find the coefficient of $x^6$ in $(2+2x+2x^2+2x^3+2x^4+x^5)^5$

I did this with a change of variables:

• $a = 2$
• $b = 2x$
• $ c = 2x^2$
• $ d = 2x^3$
• $e = 2x^4$
• $f = x^5$

And then I found out the ways I could get $x^6$ in the expansion of $(a+b+c+d+e+f)^5$

1. $a^3bf \to \frac{5!}{3!}=20$
2. $a^2c^3 \to \frac{5!}{2!3!}=10$
3. $a^3d^2 \to \frac{5!}{2!3!}=10$
4. $a^3ce \to \frac{5!}{3!}=20$
5. $a^2bcd \to \frac{5!}{2!}=60$
6. $a^2b^2e \to \frac{5!}{2!2!}=30$

Undoing the change of variables:

1. $(2)^3(2x)(x^5)=16x^5=16(20)=320$
2. $(2)^2(2x^2)^3=4(8)x^5=32(10)=320$
3. $(2)^3(2x^3)^2 = 8(4)x^5 = 32(10) = 320$
4. $(2)^3(2x^2)(2x^4) = 32(20) = 640$
5. $(2)^2(2x)(2x^2)(2x^3) = 32(60)= 1920$
6. $(2)^2(2x)^2(2x^4) = 32(30) = 960$

So the coefficient of $x^6$ would be $320+320+320+640+1920+960=4480$, but the correct answer is $6240$. Is there an easier way to do this problem?

Answer 1

Inside your list, you are missing

7. $ab^3d \to \frac{5!}{3!}=20$
8. $ab^2c^2 \to \frac{5!}{2!2!}=30$
9. $b^4c \to \frac{5!}{4!}=5$

which after the change of variables gives you

7. $(2)(2x)^3(2x^3)=32(20)=640$
8. $(2)(2x)^2(2x^2)^2=32(30)= 960$
9. $(2x)^4(2x^2)=32(5)=160$

and therefore $4480+640+960+160=6240$.

Answer 2

The special form of the polynomial makes it possible to use simple facts about the geometric series ${1\over1-x}=1+x+x^2+x^3+\cdots$ and its derivatives. We can also ignore any powers of $x$ greater than $6$, which we'll indicate here with congruence mod $x^7$ notation.

To begin, we have

$$\begin{align} 2+2x+2x^2+2x^3+2x^4+x^5 &=2(1+x+x^2+x^3+x^4+x^5)-x^5\\ &={2(1-x^6)\over1-x}-{x^5(1-x)\over1-x}\\ &={2-x^5(1+x)\over1-x} \end{align}$$

so

$$\begin{align} (2+2x+2x^2+2x^3+2x^4+x^5)^5 &\equiv\left(2-x^5(1+x)\over1-x \right)^5\\ &\equiv{2^5-5\cdot2^4x^5(1+x)\over(1-x)^5}\\ &\equiv{32-80x^5-80x^6\over(1-x)^5}\mod x^7 \end{align}$$

Now

$$\begin{align} {1\over(1-x)^5} &={1\over24}\left(1\over1-x \right)''''\\ &={1\over24}(1+x+x^2+x^3+\cdots)''''\\ &={1\over24}(x^4+x^5+x^6+x^7+\cdots)''''\quad\text{(since }(1+x+x^2+x^3)''''=0)\\ &={1\over24}(4\cdot3\cdot2\cdot1+5\cdot4\cdot3\cdot2x+6\cdot5\cdot4\cdot3x^2+7\cdot6\cdot5\cdot4x^3+\cdots)\\ &=1+5x+15x^2+35x^3+70x^4+126x^5+210x^6+\cdots \end{align}$$

so the coefficient of $x^6$ in $(2+2x+2x^2+2x^3+2x^4+x^5)^5$ is

$$32\cdot210-80\cdot5-80\cdot1=6240$$

Answer 3

Hint

$$\left(\sum_{i=1}^{n} x_i\right)^m=\sum_{i_1,\cdots i_n\\i_1+\cdots+ i_n=m}{m!\over i_1!\cdots i_n!}x_1^{i_1}\cdots x_n^{i_n}$$

Answer 4

One possible way is to work in $\Bbb Q[x]/(x^7)=\Bbb Q[[x]]/(x^7)$ (i.e. in the ring of polynomials, respectively the ring of power series in $x$, taken modulo $O(x^7)$), and i will factor $1/2^5$ first, so we compute: $$ \begin{aligned} &\left(1+x+x^2+x^3+x^4+\frac 12x^5+O(x^7)\right)^5 \\ &\qquad = \left(\frac {1-x^5}{1-x}+\frac 12x^5+O(x^7)\right)^5 \\ &\qquad = \binom 50 \left(\frac {1-x^5}{1-x}\right)^5 +\binom 51\left(\frac {1-x^5}{1-x}\right)^4\cdot\frac 12x^5+O(x^7) \\ &\qquad = \frac {(1-x^5)^5}{(1-x)^5} +5\frac {(1-x^5)^4}{(1-x)^4}\cdot\frac 12x^5+O(x^7) \\ &\qquad = (1-5x^5) \left( 1-\binom{-5}1 x +\binom{-5}2 x^2 -\binom{-5}3 x^3 +\binom{-5}4 x^4 -\binom{-5}1 x^5 +\binom{-5}6 x^6 \right) \\ &\qquad\qquad +5\cdot(1+O(x^2)) \cdot \left( 1-\binom{-4}1 x +O( x^2) \right) \frac 12x^5 \\ &\qquad\qquad\qquad+O(x^7) \end{aligned} $$ and the coefficient in $x^6$ can be extracted as $$ \binom{-5}6+5\binom{-5}1 -\frac 52\binom{-4}1 =210-25+10=195\ . $$ Recall that we have to multiply with $2^5=32$ to get the final answer $$ 195\cdot 32= 6240\ . $$