Coefficient of $x^{18}$ in $(1+x+x^2+......x^9)^3$ .

Question

We have to find coefficientof $x^{18}$ in $(1+x+x^2+......x^9)^3$ .

I wrote it as $\frac{x^9-1}{x-1}$

after that got stuck .

In the source I could not understand anything .

Answer 1

$$1+x+x^2+\cdots+x^9=\frac{1-x^{10}}{1-x}=(1-x^{10})(1-x)^{-1}$$

so

$$(1+x+x^2+\cdots+x^9)^3=\left(\frac{1-x^{10}}{1-x}\right)^3=(1-x^{10})^3(1-x)^{-3}=(1-3x^{10}+\cdots)(1+3x+6x^2+10x^3+\cdots)$$

Now collect coefficient of $x^{18}$ using $n$th term of triangular numbers $1,3,6,10,...$ as $\frac{n^2+n}{2}$

we get required answer as $$\frac{19^2+19}{2}-3 \times \frac{9^2+9}{2}=55$$

Answer 2

Since you have rewrote it as $((x^{10} - 1)(x-1)^{-1})^3$ just use the binomial theorem to expand $(x^{10}-1)^3$ and $(x-1)^{-3}$ separately using the binomial theorem. $$(x^{10} - 1)^3 = x^{30} - 3x^{20} + 3x^{10} -1$$ $$(x-1)^{-3} = (-1)^{-3} + \frac{(-3)!}{(-3 -1)! (1)!}x(-1)^{-4} + ... +\frac{(-3!)}{(-3-n)!(n)!}x^n(-1)^{-3-n}... $$

Now to get the coefficient of $x^{18}$ we look at what terms of $(x-1)^{-3}$ multiply with the terms of $(x^{10}-1)^{3}$ to give you $x^{18}$

$$ 3x^{10}\frac{(-3)!}{(-3-8)!(8!)}x^8(-1)^{-3-8} - \frac{(-3)!}{(-3-18)!(18)!}x^{18}(-1)^{-3-18}$$

Now just compute that and you will find your coefficient

Answer 3

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

We obtain \begin{align*} [x^{18}]&(1+x+x^2+\cdots+x^9)^3\\ &=[x^{18}]\left(\frac{1-x^{10}}{1-x}\right)^3\tag{1}\\ &=[x^{18}](1-x^{10})^3\cdot\frac{1}{(1-x)^3}\\ &=[x^{18}](1-3x^{10}+3x^{20}-x^{30})\sum_{n=0}^\infty\binom{-3}{n}(-x)^n\tag{2}\\ &=[x^{18}](1-3x^{10})\sum_{n=0}^\infty\binom{n+2}{n}x^n\tag{3}\\ &=\left([x^{18}]-3[x^{8}]\right)\sum_{n=0}^\infty\binom{n+2}{2}x^n\tag{4}\\ &=\binom{20}{2}-3\binom{10}{2}\tag{5}\\ &=55 \end{align*}

Comment:

• In (1) we apply the geometric series formula.

• In (2) we use the binomial series expansion.

• In (3) we skip terms with $x^{20}$ and $x^{30}$ since they do not contribute to $[x^{18}]$ and we apply the binomial identity $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$.

• In (4) we use the linearity of the coefficient of operator and apply the rule \begin{align*} [x^{p-q}]A(x)=[x^p]x^qA(x) \end{align*}

• In (5) we select the coefficients accordingly.