Find the coefficient of $ x^6 $in the given expansion.

Question

Find the coefficient of $x^6$ in $\left[(1+x)(1+x^2)^2(1+x^3)^3 \cdots (1+x^n)^n\right]$.

---

Expansion $$\left[\left(1+\binom11x \right)\left(1+\binom21x + \binom22x^2 \right)\left(1 +\binom31x+\binom32x^2+\binom33x^3 \right)\cdots \left(1 + \binom n1 x + \binom n2x^2 + \cdots +\binom n nx^n \right) \right]$$

I expanded it as shown above but couldn't proceed further. Any Help would be appreciated.

Answer 1

Taken mod $x^7$ (meaning we drop anything with a power of $x$ greater than $6$), we have

$$(1+x)(1+x^2)^2(1+x^3)^3\cdots\equiv(1+x)(1+2x^2+x^4)(1+3x^3+3x^6)(1+4x^4)(1+5x^5)(1+6x^6)\\ \equiv(1+x+5x^5+5x^6)(1+2x^2+5x^4+8x^6)(1+3x^3+9x^6)\\ $$

where in the second step we've paired $(1+x)$ with $1+5x^5$, $(1+2x^2+x^4)$ with $(1+4x^4)$, and $(1+3x^3+3x^6)$ with $(1+x^6)$. The coefficient of $x^6$ in the resulting expansion is

$$5+8+9+1\cdot2\cdot3=28$$

(where the $1\cdot2\cdot3$ comes from the product $x\cdot2x^2\cdot3x^3$).

Answer 2

Here we give a supplement regarding the cases $1\leq n\leq 5$. It is convenient to use the coefficient of operator $[x^k]$ to denote the coefficient of $x^k$ in a series.

We obtain

Cases $n=1,2$: \begin{align*} [x^6](1+x)=[x^6](1+x)(1+x^2)^2\color{blue}{=0} \end{align*}

Case $n=3$:

\begin{align*} \color{blue}{[x^6]}&\color{blue}{(1+x^3)^3(1+x^2)^2(1+x)}\\ &=[x^6](1+3x^3+3x^6)(1+2x^2+x^4)(1+x)\tag{1}\\ &=\left([x^6]+3[x^3]+3[x^0]\right)(1+2x^2+x^4)(1+x)\tag{2}\\ &=0+3\cdot2+3\cdot 1\\ &\,\,\color{blue}{=9} \end{align*}

Case $n=4$:

\begin{align*} \color{blue}{[x^6]}&\color{blue}{(1+x^4)^4(1+x^3)^3(1+x^2)^2(1+x)}\\ &=[x^6](1+4x^4)(1+3x^3+3x^6)(1+2x^2+x^4)(1+x)\\ &=[x^6](1+3x^3+4x^4+3x^6)(1+2x^2+x^4)(1+x)\tag{3}\\ &=\left([x^6]+3[x^3]+4[x^2]+3[x^0]\right)(1+2x^2+x^4)(1+x)\\ &=0 + 3\cdot 2+4\cdot 2+3\cdot 1\\ &\,\,\color{blue}{=17} \end{align*}

Case $n=5$:

\begin{align*} \color{blue}{[x^6]}&\color{blue}{(1+x^5)^5(1+x^4)^4(1+x^3)^3(1+x^2)^2(1+x)}\\ &=[x^6](1+5x^5)(1+4x^4)(1+3x^3+3x^6)(1+2x^2+x^4)(1+x)\\ &=[x^6](1+4x^4+5x^5)(1+2x^2+3x^3+x^4+6x^5+3x^6)(1+x)\tag{4}\\ &=\left([x^6]+4[x^2]+5[x]\right)(1+2x^2+3x^3+x^4+6x^5+3x^6)(1+x)\\ &=(6+3)+4\cdot 2+5\cdot 1\\ &\,\,\color{blue}{=22} \end{align*}

Comment:

• In (1) we expand the terms skipping summands with powers of $x$ greater than $6$.

• In (2) we use the rule $[x^{p-q}]A(x)=[x^p]x^qA(x)$.

• In (3) we multiply out the left-most two terms ignoring summands which do not contribute to $[x^6]$.

• In (4) we multiply out the left-most two terms and the next two terms ignoring summands which do not contribute to $[x^6]$.