Expansion of $(1+px+qx^2)^8\equiv 1+8x+52x^2+kx^3$...

Question

$(1+px+qx^2)^8 \equiv 1+8x+52x^2+kx^3...$

Given that expression, calculate the values of p, q, and k.

I'm guessing this is a trinomial that needs to be turned into a binomial, but how would you do it? I have no idea.

Answer 1

It can be done without formulae, really, reasoning combinatorially.

The left hand side is just $$(1+px+qx^2)(1+px+qx^2)(1+px+qx^2)(1+px+qx^2)(1+px+qx^2)(1+px+qx^2)(1+px+qx^2)(1+px+qx^2)$$

It's clear the only constant term we get by multiplying out is the $1$, which we indeed have on the right hand side.

The only way to get an $x$ is to have one $px$ from one of the terms and multiply it with 7 ones, from the remaining terms. As we have 8 terms, we get $8\cdot px = 8px$ when we multiply. As this should equal $8x$ we get that $p = 1$.

How to get terms $x^2$? We can pick one $qx^2$, from one of the terms and $1$ from the remaining one. So we we have $8qx^2$ from those terms. But we can also pick $px$'s from two different terms (and $1$ in the remaining $6$) and this can be done in ${8 \choose 2} = 28$ ways. So we also have $28p^2x^2 = 28x^2$ (as $p=1$) from those. So in total we have $8q + 28$ as coefficient for $x^2$, and this should equal $52$, so $8q = 24$, hence $q=3$.

Finally: the coefficient of $x^3$ equals all the ways we can pick three $px$ terms (which is ${8 \choose 3} \cdot p^3$), plus all the ways we can pick one $qx^2$ term and one $px$ term, which equals $8\cdot 7\cdot p \cdot q$. The sum of these equals $k$...