Find the coefficient of $x^{20}$ in $(x^{1}+⋯+x^{6} )^{10}$

Question

I'm trying to find the coefficient of $x^{20}$ in $$(x^{1}+⋯+x^{6} )^{10}$$

So I did this : $$\frac {1-x^{m+1}} {1-x} = 1+x+x^2+⋯+x^{m}$$

$$(x^1+⋯+x^6 )=x(1+x+⋯+x^5 ) = \frac {x(1-x^6 )} {1-x} = \frac {x-x^7} {1-x}$$

$$(x^1+⋯+x^6 )^{10} =\left(\dfrac {x-x^7} {1-x}\right)^{10}$$

But what do I do from here ? any hints ?

Thanks

Answer 1

The coefficient of $x^{10}$ in $(1+ x + \ldots + x^5)^{10}$ is equal to the number of integers $0 \leq x_i \leq 5 $ such that $\sum_{i=1}^{10} x_i= 10$.

We apply the Principle of Inclusion and exclusion, to deal with the restriction of $x_i \leq 5$.

If the only restriction is $0 \leq x_i$ then there are ${10 + 9 \choose 9 } $ solutions by the bars and stars method (sum of 10 non-negative integers is 10).

If $x_1 \geq 6$, then we substitute $x_1 = 6 + x_1 ^*$, and there are ${4 + 9 \choose 9}$ solutions by the stars and bars method (sum of 10 non-negative integers is 4).

Observe that we can't have 2 terms which are more than $6$.

Hence, by PIE, the coefficient is ${ 19 \choose 9} - 10 { 13 \choose 9}$, which is 85228.

Answer 2

You can factor out $x^{10}$ so that your goal becomes to find the $x^{10}$ coefficient in $(1 + x + .. + x^5)^{10}$. By the Multinomial Theorem one has $$(1 + x + .. + x^5)^{10} = \sum_{k_1 + ... + k_6 = 10} \frac{10!}{k_1!k_2!k_3!k_4! k_5!k_6} x^{k_2 + 2k_3 + 3k_4 + 4k_5 + 5k_6}$$ Here the sum is over nonnegative integers. So what you need to do is find all $(k_1,...,k_6)$ such that $k_2 + 2k_3 + 3k_4 + 4k_5 + 5k_6 = 10$, or equivalently all $(k_2,...,k_6)$ such that such that $k_2 + 2k_3 + 3k_4 + 4k_5 + 5k_6 \leq 10$, then add up the corresponding coefficients.

Answer 3

In fact, we need to find coefficient of $x^{10}$ in $(1+x+x^2+\cdots+x^5)^{10}$.

Easy way/trick to check this coefficient $-$ to set $$x=10^m, $$ where $m$ is rather big. Then to calculate given expression (using Mathematica or something else).

For example: $x=1\,\,000\,000$, $$ \begin{array}{l} (x^5+x^4+\cdots+x+1)^{10} = 1\;\; 000\,001\;\; 000\,001\;\; 000\,001\;\; 000\,001\;\; 000\,001^{10} = \\ \ldots \;\;\ldots \;\;\ldots \;\ldots \;\;243\,925\;\;147\,940\;\;\color{red}{085\,228}\;\; \\ 046\,420\;\;023\,760\;\;011\,340\;\;004\,995\;\;002\,002 \\ 000\,715\;\;000\,220\;\;000\,055\;\;000\,010\;\;000\,001. \end{array} $$

For $x=1\;\;0000\,0000$, $$ \begin{array}{l} (x^5+x^4+\cdots+x+1)^{10} = 1\;\; 0000\,0001\;\; 0000\,0001\;\; 0000\,0001\;\; 0000\,0001\;\; 0000\,0001^{10} = \\ \ldots \;\;\ldots \;\;\ldots \;\;\ldots \;\ldots \;\;\; 0024\,3925\;\;0014\,7940\;\;\color{red}{0008\,5228} \\ 0004\,6420\;\;0002\,3760\;\;0001\,1340\;\;0000\,4995\;\;0000\,2002\\ 0000\,0715\;\;0000\,0220\;\;0000\,0055\;\;0000\,0010\;\;0000\,0001. \end{array} $$

Answer 4

Since $(x+x^2+\cdots+x^6)^{10}=x^{10}(1+x+\cdots+x^5)^{10}$ and $1+x+\cdots+x^5=\frac{1-x^6}{1-x},$ we need to find the coefficient of $x^{10}$ in $(\frac{1-x^6}{1-x})^{10}=(1-x^6)^{10}(1-x)^{-10}.$ Since $(1-x^6)^{10}(1-x)^{-10} = (1-10x^6+45x^{12}+\cdots) \sum_{m=0}^{\infty}\binom{m+9}{9}x^{m},$ the coefficient of $x^{10}$ will be $\binom{19}{9}-10\binom{13}{9}. $