# Finding coefficient of $x^{15}$

$$(x+x^2+x^3+x^4+x^5)\cdot (x^2+x^3+x^4+…)^5$$

I have done a little :

$$x(1 + x+x^2+x^3+x^4)\cdot x^{10}(1 + x^2+x^3+…)^5$$

By generating functions:

\begin{align}&x^{11}\cdot\frac{1 - x^5}{1-x}\cdot\frac{1}{(1-x)^5}\\[1ex] \implies &x^{11}(1 - x^5)\cdot\frac{1}{(1-x)^6}\\[1.5ex] \implies &x^{11}(1 - x^5) \cdot\sum_{n=0}^{\infty}\binom{n+5}{5} x^n\\ \end{align}

How am I supposed to find $x^{15}$or any other one like $x^{18} , x^{19}$ Any clues? Thanks.

• Now think about multiplying out the product you created. You have an $x^[11}$ and either a $1$ or an $x^5$, what do you need from the 3rd term of the product in order to get $x^{15}$? – sharding4 May 21 '17 at 22:17
• @sharding4 I need $x^4$ ? – Johnathan May 21 '17 at 22:20
• Do you see any way to get an $x^4$? – sharding4 May 21 '17 at 22:22
• Check your rules for exponents. $x^{11} \cdot (-x^5)$ already has exponent greater than $15$, so that you can't get an $x^{15}$ that way. The $x^{11} \cdot 1 \cdot \binom {9}{4} x^4$ is correct. – sharding4 May 21 '17 at 22:31
• @sharding4 Hey , thanks . I just wana clarify my understanding : Suppose I had to find coefficient of $x^{19}$ for the same expansion : So the end product would be $(x^{11} - x^{16})$ .So both are fine and need 8 and 3. So would the answer be (8+5,8) - (3+5,3) ? – Johnathan May 21 '17 at 23:44

You are most of the way home with what you have done already, the last step involves manipulating the summation by multiplying through by $x^{11}(1-x^5)=x^{11}-x^{16}$ like so

\begin{align}(x^{11}-x^{16})\sum_{n=0}^{\infty}\binom{n+5}{5}x^n&=\sum_{n=0}^{\infty}\binom{n+5}{5}(x^{n+11}-x^{n+16})\\&=\sum_{n=0}^{\infty}\binom{n+5}{5}x^{n+11}-\sum_{n=0}^{\infty}\binom{n+5}{5}x^{n+16}\end{align}

relabelling summation indices $n\rightarrow n-11$ and $n\rightarrow n-16$ for the first and second summations respectively gives

$$\sum_{n=11}^{\infty}\binom{n-6}{5}x^{n}-\sum_{n=16}^{\infty}\binom{n-11}{5}x^{n}$$

but if we define $\binom{a}{b}=0$ for $a\lt b$ such that $a,b\in \mathbb{Z}$ then we can write this as

$$\sum_{n=0}^{\infty}\binom{n-6}{5}x^{n}-\sum_{n=0}^{\infty}\binom{n-11}{5}x^{n}$$

or simply

$$\sum_{n=0}^{\infty}\left(\binom{n-6}{5}-\binom{n-11}{5}\right)x^n$$

hence your coefficients $c_n$ in

$$\frac{x^{11}-x^{16}}{(1-x)^6}=\sum_{n=0}^{\infty}c_nx^n$$

are

$$c_n=\binom{n-6}{5}-\binom{n-11}{5}\tag{Answer}$$

e.g. To find the coefficient $x^{15}$ plug $n=15$ into that formula.

• Thankyou Sir. This helped a lot :D – Johnathan May 22 '17 at 1:31
• You are most welcome @Johnathan :) – N. Shales May 22 '17 at 1:44

It is convenient to use the coefficient of operator $[x^k]$ to denote the coefficient of $x^k$ in a series.

We obtain \begin{align*} \color{blue}{[x^{15}]}&\color{blue}{x^{11}(1-x^5)\sum_{n=0}^\infty \binom{n+5}{5}x^n}\\ &=[x^4](1-x^5)\sum_{n=0}^\infty\binom{n+5}{5}x^n\tag{1}\\ &=[x^4]\sum_{n=0}^\infty\binom{n+5}{5}x^n\tag{2}\\ &=\binom{9}{5}\tag{3}\\ &\color{blue}{=126} \end{align*}

Comment:

• In (1) we apply the rule $[x^{p-q}]A(x)=[x^p]x^qA(x)$.

• In (2) we note the term $x^5$ does not contribute to the coefficient of $x^4$ and can be skipped.

• In (3) we select the coefficient of $x^4$.

In general we obtain for $k\geq 0$: \begin{align*} \color{blue}{[x^{k}]}&\color{blue}{x^{11}(1-x^5)\sum_{n=0}^\infty \binom{n+5}{5}x^n}\\ &=[x^{k-11}](1-x^5)\sum_{n=0}^\infty\binom{n+5}{5}x^n\tag{4}\\ &=\left([x^{k-11}]-[x^{k-16}]\right)\sum_{n=0}^\infty\binom{n+5}{5}x^n\tag{5}\\ &\color{blue}{=\binom{k-6}{5}-\binom{k-11}{5}}\tag{6}\\ \end{align*}

Comment:

• In (4) we apply the same rule as in (1).

• In (5) we use the linearity of the coefficient of operator and apply the same rule as in (1).

• In (6) we select the coefficients accordingly. Note that here we set $\binom{n}{k}=0$ if $k>n$ or $n<0$.