How may one solve problems over expressions like $(2+px)^6$ without the binomial theorem?

Question

A friend of mine posed a problem on a mathematics discord server.

The coefficient of the $x^2$ term in the expansion of $(2+px)^6$ is $60$. Find the value of the positive constant $p$.

I immediately thought of employing the binomial theorem, as what was required, actually. But, I decided to do another method, which led me to the question of how we can solve such problems without using the binomial theorem.

How can we solve this question without the binomial theorem or newton's method?

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My Attempt:

Lemma: $$(x+a)(x+b)(x+c)=x^3+(a+b+c)x^2+(ab+bc+ac)x+abc.$$

Note that $$(2+px)^6=(2+px)^2(2+px)^2(2+px)^2$$ $$=((px)^2+4px+4)((px)^2+4px+4)((px)^2+4px+4)$$ Therefore, in the lemma, substitute $x\mapsto (px)^2$ and $a,b,c\mapsto 4px+4$. It follows $$(2+px)^6=(px)^6+12(px+1)(px)^4+48(px+1)^2(px)^2+64(px+1)^3.$$ We can ignore the first two terms since they contain no strict $x^2$ coefficient. Thus, upon expanding $$48(px+1)^2(px)^2+64(px+1)^3$$ it follows the coefficient of $x^2$ is $240p^2$. $$240p^2=60\tag{$p>0$}$$ $\therefore p=\frac 12$.

Answer 1

This implicitly uses the Binomial Theorem (edit: no it does not), and calculus, but anyway...

Let $$p(x)=(2+px)^6=a_0+a_1x+60x^2+\mathcal{O}(x^3).$$

Differentiate twice:

$$\begin{align} p'(x)&=6(2+px)^5\cdot p=a_1+60\cdot 2x+\mathcal{O}(x^2) \\ \Rightarrow p''(x)&=6p\cdot 5\cdot(2+px)^4\cdot p=120+\mathcal{O}(x) \end{align}$$

Evaluate at $x=0$:

$$30p^2\cdot 2^4=120\Rightarrow p^2=\frac{1}{4}\Rightarrow p\underset{p>0}{=}+\sqrt{\frac14}=\frac12.$$

Answer 2

Here are three different methods. The first one is recommended, the others are just for fun and curiosity.

Combinatorial approach:

We use the coefficient of operator $[x^n]$ to denote the coefficient of $x^n$ of a series. Here we are looking for a positive solution $p$ of \begin{align*} 60=[x^2](2+px)^6\tag{1} \end{align*} Multiplying $(2+px)^6$ out we get a contribution to $x^2$ if and only if we select from two factors $2+px$ the term $px$ giving us $p^2x^2$ and taking from the other $4$ factors $2$ giving us $2^4=16$. Since we can select two factors from $(2+px)^6$ in $\binom{6}{2}$ ways, we obtain \begin{align*} \color{blue}{60}=[x^2](2+px)^6=\binom{6}{2}p^22^4=15p^2\cdot 16\color{blue}{=240p^2} \end{align*} from which $p=\frac{1}{2}$ follows.

Algebraic approach:

A somewhat cumbersome but simple approach is to iteratively work through the linear factors $2+px$.

We obtain \begin{align*} \color{blue}{60}&=[x^2](2+px)^6\\ &=[x^2](2+px)(2+px)^5\\ &=\left(2[x^2]+p[x^1]\right)(2+px)(2+px)^4\tag{1}\\ &=\left(4[x^2]+4p[x^1]+p^2[x^0]\right)(2+px)(2+px)^3\tag{2}\\ &=\left(8[x^2]+12p[x^1]+6p^2[x^0]\right)(2+px)(2+px)^2\\ &=\left(16[x^2]+32p[x^1]+24p^2[x^0]\right)(2+px)(2+px)^1\\ &=\left(32[x^2]+80p[x^1]+80p^2[x^0]\right)(2+px)\\ &=160p^2+80p^2\tag{3}\\ &\,\,\color{blue}{=240p^2} \end{align*} from which $p=\frac{1}{2}$ follows.

Comment:

• In (1) we use the linearity of the coefficient of operator and apply the rule $[x^p]x^qA(x)=[x^{p-q}]A(x)$.

• In (2) and the following lines we determine the coefficients according to the rules from (1), i.e. \begin{align*} &\left(a[x^2]+b[x^1]+c[x^0]\right)(2+px)=2a[x^2]+(2b+ap)[x^1]+(2c+bp)[x^0] \end{align*}

• In (3) we select the coefficients accordingly.

Complex analytic approach:

This variant shouldn't be taken too serious. It's just for fun and in fact based on the first method. We recall the residue theorem which tells us that integrating along a circle with radius one around the origin we have \begin{align*} [x^2](2+px)^6=\frac{1}{2\pi i}\oint_{|x|=1}\frac{(2+px)^6}{x^3}\,dx \end{align*}

We obtain \begin{align*} \color{blue}{60}&=[x^2](2+px)^6\\ &=\frac{1}{2\pi i}\oint_{|x|=1}\frac{(2+px)^6}{x^3}\,dx\\ &=\frac{1}{2\pi i}\int_{0}^{2\pi}\frac{\left(2+pe^{it}\right)^6}{e^{3it}}\,ie^{it}\,dt\tag{2}\\ &=\frac{1}{2\pi}\int_{0}^{2\pi}\left(2+pe^{it}\right)^6e^{-2it}\,dt\tag{3}\\ &=\frac{1}{2\pi}\int_{0}^{2\pi}\binom{6}{2}2^4p^2\,dt\tag{4}\\ &=\frac{1}{2\pi}\binom{6}{2}2^4p^2\int_{0}^{2\pi}\,dt\\ &\,\,\color{blue}{=240p^2} \end{align*} from which $p=\frac{1}{2}$ follows.

Comment:

• In (2) we use the substitution $x=e^{it}, dx=ie^{it}dt$.

• In (3) we note $\int_{0}^{2\pi}e^{-kit}\,dt =0, k\in \mathbb{Z}$ which is due to Euler's identity $e^{2\pi i}=1$. So, everything vanishes in (3) besides the constant term.

• In (4) we select the constant term in fact with the same considerations we used in the first approach.

Note: Another approach might be using shifted series multisection in order to filter the wanted coefficient of $x^2$.