# How can I find the coefficient of x when the power is greater than the powers of 2 brackets using binomial expansion?

I have been given this question:

Find the coefficient of $x^{13}$ in the expansion of $(1 + 2x)^4(2 + x)^{10}$.

I know how I would find $x^4$ or lower degrees, but I am unsure how to approach this, as neither term has a $x^{13}$, and x is a prime number so it can't just be 2 terms multiplied (as neither bracket has a power of 13).

This is revision rather than homework, but hints would be appreciated.

• You have to think about which terms in the first bracket, when expanded, would have to multiply which terms in the second bracket, when expanded, in order to create the required term. Note that $x^a\times x^b=x^{a+b}$. – David Quinn Jan 11 '16 at 19:51

If you expand the two terms, the first one will give you terms from $1$ to $16x^4$. The second will give you terms from $2^{10}$ to $x^{10}$. When you multiply them, the only two ways to get $x^{13}$ is to use the $x^3$ term from the first and the $x^{10}$ term from the second or to use the $x^4$ from the first and the $x^9$ from the second. Evaluate the coefficients of each of these terms, multiply, and add.

• Although tyt's answer was first, I think this one is slightly clearer at explaining what I dp. Thankyou! – Tim Jan 11 '16 at 19:53

$(1+2x)^4$ give out terms containing $x$ with power $0,1,2,3,4$ similarly $(2+x)^{10}$ give out term containing $x$ with power $0$ to $10$ now pick up exponents from first and second such that these sum up to 13 along with coefficient. ATP, $$(10{C_0}x^{10})(4C_3(2x)^3)+2(10C_1x^9)(4C_4(2x)^4)$$ and these will give out coefficient.

Since $x^{10}\cdot x^3=x^{13}$ and $x^{9}\cdot x^4=x^{13}$, we just have to find the coefficients of $x^4,x^3$ in $(1+2x)^4$ and the coefficients of $x^{10},x^9$ in $(2+x)^{10}$. To do this you can use binomial thereom: $$x^3:{4 \choose 3}(2x)^3=32x^3$$ $$x^4:{4 \choose 0}(2x)^4=16x^4$$ $$x^{10}:{10 \choose 0}(x)^{10}=x^{10}$$ $$x^9:{10 \choose 1}2(x)^9=20x^9.$$ So we have $20x^9\cdot 16x^4+32x^3\cdot x^{10}=352x^{13}.$

There is a useful notation, the so-called coefficient of operator $[x^n]$ to denote the coefficient of $x^n$ of a polynomial or a series. With this notation you can do some kind of bookkeeping of the coefficients you need from the binomials.

We obtain \begin{align*} [x^{13}]&(1+2x)^4(2+x)^{10}\\ &=[x^{13}]\left(\sum_{j=0}^4\binom{4}{j}2^jx^j\right)(2+x)^{10}\tag{1}\\ &=\left(\sum_{j=0}^4\binom{4}{j}2^j[x^{13-j}]\right)(2+x)^{10}\tag{2}\\ &=\left(\binom{4}{3}2^3[x^{10}]+\binom{4}{4}2^4[x^9]\right)(2+x)^{10}\tag{3}\\ &=\left(32[x^{10}]+16[x^9]\right)\sum_{j=0}^{10}\binom{10}{j}2^{10-j}x^j\tag{4}\\ &=32\binom{10}{10}2^0+16\binom{10}{9}2^1\tag{5}\\ &=352 \end{align*}

Comment:

• In (1) we expand the left binomial

• In (2) we use the linearity of the coefficient of Operator and $[x^n]x^kp(x)=[x^{n-k}]p(x)$

• In (3) we see that only $j=3$ and $j=4$ provide a contribution

• In (4) we expand the other binomial expression

• In (5) we take the coefficients of $x^{10}$ and $x^9$