Find the coefficient of $x^{21}$ in $(1+x+x^2+\dots+x^{10})^4$
I tried splitting the terms inside the bracket into two parts $1+x+\dots+x^9$ and $x^{10}$, and then tried binomial theorem, but that got unreasonably hard to deal with.
Another idea that came is to write the expression inside bracket as $\smash{\dfrac{x^{11}-1}{x-1}}$, but I don't know how to proceed. Help
$$f(x)=\bigg(\frac{x^{11}-1}{x-1}\bigg)^4=\bigg(\frac{1-x^{11}}{1-x}\bigg)^4=(1-x^{11})^4(1-x)^{-4}$$
Now by binomial expansion for negative powers, we know
$$(1+y)^n=\sum_{k=0}^{∞}\binom{n}{k} y^k$$
Here $y=-x,n=-4$.
Hence $$(1-x)^{-4}=\sum_{k=0}^{∞}\binom{-4}{k} (-x)^k=1+\binom{-4}{1}(-x)^1+\binom{-4}{2}(-x)^2+...$$
And by normal binomial expansion,
$$(1-x^{11})^4=\sum_{k=0}^4 (-1)^k \binom {4}{k} (x^{11})^k$$
Hence we only get a single term with $x^{21}$ when the term with $x^{11}$ is multiplied by $x^{10}$. Coefficient of $x^{11}=-4$ and Coefficient of $x^{10}= \binom{-4}{10} =\frac{-4×-5×-6×-7×-8×-9×-10×-11×-12}{9!}=-220$.
Thus coefficient of $x^{21}=(-220) \cdot (-4) =880$.
Find the coefficient of $x^{21}$ in $(1+x+x^2+\dots+x^{10})^4$
For the coefficient of $x^{21}$, we need to sum all combinations that result in their total product being $x^{21}$
$$(1+x+x^2+\dots+x^{10})*(1+x+x^2+\dots+x^{10})*(1+x+x^2+\dots+x^{10}) *(1+x+x^2+\dots+x^{10})$$
Let $$A=\left(10^{0}+10^{1}+10^{2}+10^{3}+10^{4}+10^{5}+10^{6} + 10^{7}+10^{8}+10^{9}+10^{10} \right)$$
Then, with: $$C=(1+x+x^2+\dots+x^{10})*(1+x+x^2+\dots+x^{10})*(1+x+x^2+\dots+x^{10}) *(1+x+x^2+\dots+x^{10})$$ And: $$B=A*A=(1+x+x^2+\dots+x^{10})*(1+x+x^2+\dots+x^{10}) \\ \text{And }C=B*B$$ Then, the convolution needed to solve for $B$ from $A$ (where $B_i$ is the coefficient for the element of $B$ that multiplies $10^i$) is a simple addition of $1 \textit{ elements}$. And there is only one coefficient of $C=B*B$ that needs to be solved for, namely $C_{21}$ that is the coefficient of $10^{21}$ for $C$. Start by solving for $B$ that ranges from $10^0$ to $10^{20}$ (21 coefficients from $B_0$ to $B_{20}$ in all): $$B=1*10^0+2*10^1+3*10^2+4*10^3+5*10^4+6*10^5+7*10^6+8*10^7+9*10^8 +10*10^9+11*10^{10}+10*10^{11}+9*10^{12}+8*10^{13}+7*10^{14}+ 6*10^{15}+5*10^{16}+4*10^{17}+3*10^{18}+2*10^{19}+1*10^{20}$$ Now the convolution of $B$ and $B$ is needed, namely $C=B*B$. However, by hand, it is not necessary to calculate all coefficients of $C_k*10^{k}$. The calculation is needed only for $C_{21}*10^{21}$. There are references on digital Linear Convolution of sequences, but the concept is simple enough that it can be figured out here. Say there is a contributing factor on the left term of the convolution $B_i*10^i$ and on the right term $B_j*10^j$. The requirement for all contributing products is that $i+j=21$, so that $B_i*10^i*B_j*10^j=B_i*B_j*10^{21}$ which is by inspection correct.
On the left, $B_0$ makes zero contribution, since there is no nonzero term $B_{\left({21}-0\right)}=B_{21}=0$, and the same with $B_0$ on the right, a zero contribution. So the calculation can proceed, starting with $B_1$ on the left and finishing with $B_1$ on the right:
$$ C_{21}=B_1*B_{20}+B_2*B_{19}+B_3*B_{18}+B_4*B_{17}+B_5*B_{16} +B_6*B_{15}+B_7*B_{14}+B_8*B_{13}+B_9*B_{12}+B_{10}*B_{11} +B_{11}*B_{10}+B_{12}*B_9+B_{13}*B_8+B_{14}*B_7+B_{15}*B_6 +B_{16}*B_5+B_{17}*B_4+B_{18}*B_3+B_{19}*B_2+B_{20}*B_1 $$
This final calculation can be done by hand or with a hand calculator (here the Unix Bash shell \$(()) Command Shell Calculator so that the calculation is entirely visible):
echo $((2*1+3*2+4*3+5*4+6*5+7*6+8*7+9*8+10*9+11*10+10*11+9*10+8*9+7*8+6*7+5*6+4*5+3*4+2*3+1*2))
880
$$ \boxed{\text{The coefficient of }10^{21}\text{ is: } \text{Result of how many }10^{21}{ is: }880}$$
$$ \begin{align*} [1\text{ }1\text{ }1\text{ }1\text{ }1\text{ }1\text{ }1\text{ }1\text{ }1\text{ }1\text{ }1]* [1\text{ }1\text{ }1\text{ }1\text{ }1\text{ }1\text{ }1\text{ }1\text{ }1\text{ }1\text{ }1]* [1\text{ }1\text{ }1\text{ }1\text{ }1\text{ }1\text{ }1\text{ }1\text{ }1\text{ }1\text{ }1]* [1\text{ }1\text{ }1\text{ }1\text{ }1\text{ }1\text{ }1\text{ }1\text{ }1\text{ }1\text{ }1] \end{align*} $$
The full convolution results in 41 elements, with element 0 corresponding to the coefficient of $x^0$ and element 41 corresponding to $x^{40}$. Because of this, the coefficient of $x^{21}$ is actually the $22^{nd}$ entry in the convolution which results in the coefficient value of $880$:
$$ \left[ \text{ }1\text{ }\text{ }4\text{ }\text{ }10\text{ }\text{ }20\text{ }\text{ }35\text{ }\text{ }56\text{ }\text{ }84\text{ }\text{ }120\text{ }\text{ }165\text{ }\text{ }220\text{ }\text{ }286\text{ }\text{ }360\text{ }\text{ }439\text{ }\text{ }520\text{ }\text{ }600\text{ }\text{ }676\text{ }\text{ }745\text{ }\text{ }804\text{ }\text{ }850\text{ }\text{ }880\text{ }...\\ ...\text{ }891\text{ }\textbf{ 880}\text{ }\text{ }850\text{ }\text{ }804\text{ }\text{ }745\text{ }\text{ }676\text{ }\text{ }600\text{ }\text{ }520\text{ }\text{ }439\text{ }\text{ }360\text{ }\text{ }286\text{ }\text{ }220\text{ }\text{ }165\text{ }\text{ }120\text{ }\text{ }84\text{ }\text{ }56\text{ }\text{ }35\text{ }\text{ }20\text{ }\text{ }10\text{ }\text{ }4\text{ }\text{ }1\text{ }\right] $$
Thus:
$$ \boxed{\text{The coefficient of }10^{21}\text{ is: } \text{Result of how many }10^{21}{ is: }880}$$
Below are the Tables Showing the Term Contributions, starting from 1 to 200, next from 201 to 400, next from 401 to 600, next from 601 to 800, and next from 801 to 880. Following the procedure in the table, the coefficients for the result of $880*10^{21}$ can be counted out by hand. And the hand calculations can be shortened by noticing patterns in the below tables.
The result was calculated on the online C compiler as follows:
// Basic Ountline From: https://www.programiz.com/c-programming/online-compiler/
// Online C compiler to run C program online
#include <stdio.h>
int i; //Power in first term
int j; //Power in second term
int k; //Power in third term
int l; //Power in fourth term
int itotal; //Number of terms that has 21 total powers
int isubtotal; //sum of i, j, k, and l
int main() {
itotal=0;
for (i=0; i<=10; i++)
for (j=0; j<=10; j++)
for (k=0; k<=10; k++)
for (l=0; l<=10; l++)
{
isubtotal=i+j+k+l;
if (isubtotal ==(int) 21)
{
itotal++;
}
// printf("isubtotal = %d \n", isubtotal);
}
printf("Result of how many 10^21 is: %d \n",itotal);
return 0;
}
// Basice Ountline From: https://www.programiz.com/c-programming/online-compiler/
// Online C compiler to run C program online
#include <stdio.h>
int i; //Power in first term
int j; //Power in second term
int k; //Power in third term
int l; //Power in fourth term
int m; //Which coefficient being sought
int itotal; //Number of terms that has m total powers
int isubtotal; //sum of i, j, k, and l
int main() {
printf("$$ \n \left[ ");
for (m=0; m<41; m++)
{
for (itotal=0, i=0; i<=10; i++)
for (j=0; j<=10; j++)
for (k=0; k<=10; k++)
for (l=0; l<=10; l++)
{
isubtotal=i+j+k+l;
if (isubtotal == m)
{
itotal++;
}
}
printf("\text{ }%d\text{ }", itotal);
}
printf("\right] \n $$\n");
return 0;
}
// Basic Outline From: https://www.programiz.com/c-programming/online-compiler/
// Online C compiler to run C program online. Or it is also possible to compile with programs such as gcc.
#include <stdio.h>
int i; //Power in first term
int j; //Power in second term
int k; //Power in third term
int l; //Power in fourth term
int itotal; //Number of terms that has 21 total powers
int isubtotal; //sum of i, j, k, and l
int main() {
printf("$$\n \begin{aligned} C10^{21}&=0 \text{ } &\text{ }\\ \n");
itotal=0;
for (i=0; i<=10; i++)
for (j=0; j<=10; j++)
for (k=0; k<=10; k++)
for (l=0; l<=10; l++)
{
isubtotal=i+j+k+l;
if (isubtotal ==(int) 21)
{
itotal++;
printf("&+10^{%d}10^{%d}10^{%d}10^{%d} \text{ } & \text{Term # }%d \\ \n",i,j,k,l,itotal);
}
}
printf("\\ \n \\ \n \boxed{\text{Result of how many 10^{21} is }C\text{ is: }C=%d} \n \end{aligned}$$",itotal);
return 0;
}
$\displaystyle (1+x+...+x^{10})^4 = \sum_{0\le a,b,c,d\le 10} x^{a+b+c+d}=\sum_{k=0}^{40}\sum_{0\le a,b,c,d\le 10\\ a+b+c+d=k}n_{4}(k)x^k$
Where $n_m(k)$ is the number of ways $k$ can be written as a sum of $m$ integers that are each less or equal than $10$.
$\displaystyle \begin{array}{ccl} n_4(21)&=&card((a,b,c,d)\in[\![0,10]\!]^4| a+b+c+d=21)\\ &=& card(\cup_{a=0}^{10}((b,c,d)\in[\![0,10]\!], b+c+d=21-a)\\ &=&\sum_{a=0}^{10}n_3(21-a)\\ &=&...\\ &=&\sum_{a=0}^{10}\sum_{b=0}^{10}n_2(21-(a+b))\\ &=&\sum_{\ell=0}^{20}\sum_{a,b=0\\a+b=\ell}^{10}n_2(21-\ell)\\ &=&\sum_{\ell=0}^{20}n_2(\ell)n_2(21-\ell)\end{array}$
Now, it's easy to see that $n_2(\ell)=\ell+1$ if $\ell\in[\![0;10]\!]$ et $n_2(\ell)=21-\ell$ if $\ell\in[\![11,21]\!]$
Whence
$$\begin{array}{ccl} n_4(21)&=&\sum_{\ell=0}^{20}n_2(\ell)n_2(21-\ell)\\ &=&\sum_{\ell=0}^{10}(\ell+1)(21-(21-\ell))+\sum_{\ell=11}^{20}(21-\ell)(21-(\ell-1))\\ &=&\sum_{\ell=1}^{10}\ell(\ell+1)+\sum_{\ell=1}^{10}\ell(\ell+1)\\ &=&2\sum_{\ell=1}^{10}\ell^2+\ell\\ &=&\frac{10\times 11\times 21}{3}+10\times 11\\ &=&880\end{array}$$
Conclusion : the coefficient before $x^{21}$ is $880$.
This is a "stars and bars" problem.
How many ways are there to order 21 stars and 3 bars with no more that 10 stars between each bar.
Another equivalent formulation:
How many ways are there to divide 21 objects into 4 piles such that no pile has more than 10 objects in each pile (piles can be empty)?
The number of ways of rolling a 21 on 4 dice (if the dice have 11 sides and one of those sides is numbered 0) -- okay, this example is less intuitive.
$\frac {24!}{21!3!} = {24\choose 3}$ This is the number of ways to order 21 stars and 3 bars.
But then we must exclude the cases where there are more than $10$ stars between one of the stars. So, we overload one bucket (by putting 11 stars in a bucket) and count the ways to allocate the remaining stars. There are 4 choices for the overfilled bucket. There are 10 remaining stars after placing 11 in one bucket.
There ${13\choose 3}$ ways to allocate the remaining 10 stars and the 3 bars.
With the limitations of the problem, we do not have to worry about overfilling 2 buckets.
${24\choose 3} - 4{13\choose 3}$
Alternative approach
Consider the following tableau
1 + x + x^2 + \cdots + x^9 + x^{10} : y_1
1 + x + x^2 + \cdots + x^9 + x^{10} : y_2
1 + x + x^2 + \cdots + x^9 + x^{10} : y_3
1 + x + x^2 + \cdots + x^9 + x^{10} : y_4
Letting each of $~y_1, y_2, y_3, y_4~$ range through the $~11~$ possible terms, you will have exactly $~11^4~$ products of the form $~(y_1 \times y_2 \times y_3 \times y_4).~$ So, the question is:
Exactly how many of these $~11^4~$ products result in the value $~x^{21}~?$
Now consider the following Stars and Bars problem: enumerate the number of solutions to
Clearly, there is a bijection between the set of solutions to the bullet pointed problem above, and the subset of the $~11^4~$ products that result in $~x^{21}.~$
See this answer for a blueprint of how to combine Inclusion-Exclusion with Stars-And-Bars to attack this generic type of problem. Following the model in the linked answer, the computation is
$$\binom{24}{3} - \left[ ~4 \times \binom{13}{3} ~\right] = 880.$$
