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.
Best Answer
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 $x\cdot2x^2\cdot3x^3$).