If $(1+ax+bx^{2})^{10} = 1-30x+410x^{2}+...$ find the value of a and b then find the coefficient of $x^{19}$ in this expansion.

Question

If: $$(1+ax+bx^{2})^{10} = 1-30x+410x^{2}+...$$ find the value of a and b then find the coefficient of $x^{19}$ in this expansion.

I found $a=-3$ and $b=\frac{1}{2}$ by writing the trinomial as a binomial as follows: $$((1)+(ax+bx^{2}))^{10} = 1-30x+410x^{2}+...$$ But I have no idea how to find the coefficient of a specific power aside from expanding fully which would be an epic job in this case.

The answer in the book is $\frac{-15}{256}$.

How is this done?

Answer 1

You can use for the coefficients the triangle of Pascal which gives $$1,10,45,120,210,252,210,120,45,10,1$$

Then$$[(ax+bx^2)+1]^{10}=(ax+bx^2)^{10}+10(ax+bx^2)^9+.....+252(ax+bx^2)^5+...+1$$ From this you get

$$10a=-30\Rightarrow a=-3\\45a^2+10b=410\Rightarrow b=\frac 12$$

For the coefficient of $x^{19}$you can developpe $(ax+bx^2)^{10}$ (using the same coefficients above) which gives $10ab^9$ as coefficient of $x^{19}$. Do you have to find if $(ax+bx^2)^9$ and the others powers have a power $x^{19}$ to finish?

Answer 2

Here is a slighty different variation of the theme. It is convenient to use the coefficient of operator $[x^n]$ to denote the coefficient of $x^n$ of a polynomial.

We consider the polynomial \begin{align*} (1+ax+bx^2)^{10}=1-30x+410x^2+\cdots\tag{1} \end{align*}

$$ $$

Coefficient $[x^1]$:

We obtain \begin{align*} [x^1](1+ax+bx^2)^{10}&=[x^1]\sum_{k=0}^{10}\binom{10}{k}(ax+bx^2)^k\\ &=[x^1]\binom{10}{1}(ax+bx^2)\\ &=10a \end{align*} Since we are looking for $[x^1]$ only $k=1$ provides a contribution, resulting finally in $10a$.

Comparison with (1) gives \begin{align*} 10a&=-30\\ a&=-3\\ \end{align*}

$$ $$

Coefficient $[x^2]$:

We obtain \begin{align*} [x^2](1+ax+bx^2)^{10}&=[x^2]\sum_{k=0}^{10}\binom{10}{k}(ax+bx^2)^k\\ &=[x^2]\binom{10}{1}(ax+bx^2)+[x^2]\binom{10}{2}(ax+bx^2)^2+\\ &=10b+45a^2 \end{align*} Since we are looking for $[x^2]$ only $k=1$ and $k=2$ provide a contribution, resulting finally in $10b+45a^2$.

Comparison with (1) gives \begin{align*} 10b+45a^2&=410\\ 10b+405&=410\\ b&=\frac{1}{2} \end{align*}

We conclude the polynomial under consideration is \begin{align*} (1-3x+\frac{1}{2}x^2)^{10} \end{align*}

Coefficient $[x^{19}]$:

We obtain \begin{align*} [x^{19}](1-3x+\frac{1}{2}x^2)^{10}&=[x^{19}]\sum_{k=0}^{10}\binom{10}{k}(-3x+\frac{1}{2}x^2)^k\\ &=[x^{19}]\binom{10}{10}(-3x+\frac{1}{2}x^2)^{10}\\ &=[x^{19}]x^{10}(-3+\frac{1}{2}x)^{10}\\ &=[x^{9}](-3+\frac{1}{2}x)^{10}\tag{1}\\ &=[x^{9}]\sum_{j=0}^{10}\binom{10}{j}\left(\frac{1}{2}x\right)^j(-3)^{10-j}\tag{2}\\ &=\binom{10}{9}\left(\frac{1}{2}\right)^9(-3)^1\\ &=-\frac{15}{256} \end{align*}

Since we are looking for $[x^{19}]$ only $k=10$ provides a contribution.

In (1) we use the rule $[x^{n-k}]p(x)=[x^n]x^{k}p(x)$.

In (2) we observe that only $j=9$ provides a contribution.