In the expansion of $(ax+b)^{2000}$, where a and b are relativel prime integers, the coefficients of $x^2$ and $x^3$ are equal. What is (a, b)?

Question

I think I would use the binomial theorem here, but how would I make those two coefficients equal if they are not middle terms?

Answer 1

The binomial expansion would tell us that the $j$th coefficient (i.e. the coefficient on $x^j$) is

$$a^jb^{2000-j}\binom{2000}{j}.$$

Do this for $j=2$ and $j=3$ and equate them. What can you conclude? (Recall that $\binom{n}{r} = \frac{n!}{(n-r)!r!}$.)

Answer 2

The binomial theorem can be used to find the coefficient of any term.

$$T_{r+1}= {{n}\choose{r}}a^{n-r}b^r$$

Before starting, flip the expansion to $(b+ax)^{2000}$. It'll make things a lot easier later on.

We want the $x^2$ term to equal the $x^3$ term, so let's equate the $3^{\text{rd}}$ and $4^{\text{th}}$ terms.

\begin{align} {2000 \choose 2}(b)^{2000-2}(ax)^{2} & = {2000 \choose 3}(b)^{2000 - 3}(ax)^3 \\ \frac{2000!}{(2000-2)!2!}b^{1998}a^2x^2& =\frac{2000!}{(2000-3)!3!}b^{1997}a^3x^3 \end{align}

We can safely ignore the $x^2$ and $x^3$ as we are only dealing with coefficients. Now remains the trivial task of simplifying the equation.

\begin{align} \frac{2000!}{(2000-2)!2!}b^{1998}a^2& =\frac{2000!}{(2000-3)!3!}b^{1997}a^3 \\ \frac{2000!}{1998!2!}b^{1998}a^2& = \frac{2000!}{1997!3!}b^{1997}a^3 \\ \frac{2000\cdot 1999}{2}\cdot b^{1998}a^2& = \frac{2000\cdot 1999\cdot 1998}{3\cdot2}\cdot b^{1997}a^3 \\ \frac{1}{2}a^2b^{1998}& = 333a^3b^{1997}\\ a^2b^{1998} & = 666a^3b^{1997} \\ b & = 666a \\ \end{align}

Now, if $b=666a$ then they obviously there is the solution $a=1$ and $b=666$. Any other solution will lead to $a$ and $b$ not being coprime. Hence the only solution is $(a,b)=(1,666)$