Finding coefficient of $x^{100}$

Question

What will be solution of this function for coefficient of $x^{100}$?

$$\displaystyle\frac{1}{\left ( 1-x^{10} \right )(1-x^{20})(1-x^{50})}$$

My solution:

$[x^{100}] \ \ \ \ \ \ \ \ \  \ $$\displaystyle \frac{1}{\left ( 1-x^{10} \right )(1-x^{20})(1-x^{50})}$

$[x^{100}] \ \ \ \ \ \ \ \ \  \ $ $(1-x^{10})^{-1}$$(1-x^{20})^{-1}$$(1-x^{50})^{-1}$

$\begin{aligned}\displaystyle &\binom{-n}{k} = (-1)^r\binom{n + k - 1}{k}\\ &=\displaystyle \color{red} {(-1)^{10}\binom{-1}{10}\binom{-1}{0}\binom{-1}{0} + (-1)^5 \binom{-1}{0}\binom{-1}{5}\binom{-1}{0} + (-1)^2\binom{-1}{0}\binom{-1}{0}\binom{-1}{2}} \\ & + \color{blue}{ (-1)^8 (-1)^1 \binom{-1}{8}\binom{-1}{1}\binom{-1}{0} + (-1)^6 (-1)^2\binom{-1}{6}\binom{-1}{2}\binom{-1}{0} + (-1)^4 (-1)^3 \binom{-1}{4}\binom{-1}{3}\binom{-1}{0} + (-1)^2 (-1)^4\binom{-1}{2}\binom{-1}{4}\binom{-1}{0} }\\ & + \color{Orange}{ (-1)^5 (-1)^1 \binom{-1}{5}\binom{-1}{0}\binom{-1}{1} } \\ & + \color{maroon}{(-1)^3(-1)^1(-1)^1\binom{-1}{3}\binom{-1}{1}\binom{-1}{1} + (-1)^1 (-1)^2(-1)^3\binom{-1}{1}\binom{-1}{2}\binom{-1}{1} }\end{aligned} $

$ 1 – 1 + 1 + 1 – 1 + 1 – 1 – 1 + 1 – 1 = 0$

So coefficient of $x^{100}$ is $0$

There are many terms so to differentiate among them distinct colors are used

$\color{red}{\text{Red}}$ - Selecting $x$ from any one of $(1-x^{10})^{-1}$$(1-x^{20})^{-1}$$(1-x^{50})^{-1}$

$\color{Blue}{\text{Blue}}$ - Selecting from both $(1-x^{10})^{-1}$$(1-x^{20})^{-1}$

$\color{Orange}{\text{Orange}}$ - Selecting from both $(1-x^{10})^{-1}$$(1-x^{50})^{-1}$

$\color{maroon}{\text{Maroon}}$ - Selecting from all $(1-x^{10})^{-1}$$(1-x^{20})^{-1}$$(1-x^{50})^{-1}$

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But answer is $10$, is anything wrong with my solution?

Answer 1

Hint:

Set $y=x^{10},$

Now for $|y|<1, (1-y)^{-1}=1+y+y^2+\cdots$

we need the coefficient of $y^{10}$ in $$(1-y^5)^{-1}(1-y^2)^{-1}(1-y)^{-1}$$

$$=(1+y^5+y^{10}+\cdots)(1-y^2)^{-1}(1-y)^{-1}$$

So, the coefficient of $y^{10}$ in $(1-y^2)^{-1}(1-y)^{-1}$(which is $6$ due to $1\cdot y^{10},y^2\cdot y^8,y^4\cdot y^6,y^6\cdot y^4,y^8\cdot y^2,,y^{10}\cdot y^0)$

$+$ the coefficient of $y^{10-5}$ in $(1-y^2)^{-1}(1-y)^{-1}$ (which is $3,$ how ? )

$+$ the coefficient of $y^{10-10}$ in $(1-y^2)^{-1}(1-y)^{-1}$ (which is $1$ )

Answer 2

The problem is equivalent to that of how many ways of making one pound out of ten-pence, twenty-pence and fifty-pence pieces, an just writing out all possibilities rapidly gives ten of them.

I'll try to explain in monochrome. For any $t$, $$\frac1{1-t}=\sum_{n=0}^\infty t^n$$ at least if $t$ is near zero. Then \begin{align} \frac1{(1-x^{10})(1-x^{20})(1-x^{50})} &=(1+x^{10}+x^{20}+x^{30}+\cdots)\\ &\times (1+x^{20}+x^{40}+x^{60}+\cdots)\\ &\times (1+x^{50}+x^{100}+x^{150}+\cdots) \end{align} and we see the coefficients of all powers of $x^{10}$ in this expansion are all positive. More formally, \begin{align} \frac1{(1-x^{10})(1-x^{20})(1-x^{50})} &=\sum_{a=0}^\infty x^{10a}\sum_{b=0}^\infty x^{20b}\sum_{c=0}^\infty x^{50c}\\ &=\sum_{a,b,c=0}^\infty x^{10a+20b+50c} \end{align} so the coefficient of $x^{100}$ is the number of solutions to $10a+20b+50c=100$ in non-negative integers.

Answer 3

Your mistake is in calculating $\binom{-1}k$. You had $$ (-1)^{10}\binom{-1}{10}+(-1)^5\binom{-1}{5}+\dots=(-1)^{10}\cdot 1+(-1)^5\cdot \color{red}1+\dots $$ The one in red is a mistake. $\binom{-1}{5}=\frac{(-1)(-2)\cdots(-5)}{5!}=(-1)^5=-1$. In fact, your summation should just be a summation of ten ones, because each summand is a product of numbers of the form $$ (-1)^k\binom{-1}{k}=(-1)^k(-1)^k=(-1)^{2k}=1 $$ The answer is then $1+1+1+1+1+1+1+1+1+1=10$.