In expansion of $(1+x+x^{2}+x^{3}+....+x^{27})(1+x+x^{2}+x^{3}+....x^{14})^{2}$ . Find the coefficient of $x^{28}$

Question

I am not able to apply binomial theorem here

$(1+x+x^{2}+x^{3}+....+x^{27})(1+x+x^{2}+x^{3}+....x^{14})^{2}$ Please help me to find the coefficient of$ x^{28}$ Any help will be appreciated.

Answer 1

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

In order to determine the coefficient of $x^{28}$ we obtain \begin{align*} \color{blue}{[x^{28}]}&\color{blue}{(1+x+\cdots+x^{27})(1+x+\cdots+x^{14})^2}\\ &=[x^{28}]\frac{1-x^{28}}{1-x}\cdot\frac{(1-x^{15})^2}{(1-x)^2}\tag{1}\\ &=[x^{28}]\frac{(1-x^{28})(1-2x^{15}+x^{30})}{(1-x)^3}\\ &=[x^{28}](1-2x^{15}-x^{28})\sum_{j=0}^\infty\binom{-3}{j}(-x)^j\tag{2}\\ &=\left([x^{28}]-2[x^{13}]-[x^0]\right)\sum_{j=0}^\infty\binom{j+2}{j}x^j\tag{3}\\ &=\binom{30}{2}-2\binom{15}{2}-\binom{2}{0}\tag{4}\\ &=435-210-1\\ &\color{blue}{\,\,=224} \end{align*}

Comment:

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

• In (2) we expand the numerator $(1-x^{28})(1-x^{15})^2$ and skip all terms with exponent greater $28$ since they don't contribute to the coefficient of $x^{28}$. We also apply the binomial series expansion

• In (3) we use the linearity of the coefficient of operator and apply the rule $[x^{p-q}]A(x)=[x^p]x^qA(x)$. We also apply the binomial identity \begin{align*} \binom{-p}{q}=\binom{p+q-1}{q}(-1)^q \end{align*}

• In (4) we select the coefficients accordingly.

Answer 2

\begin{eqnarray*} (1+x+x^{2}+x^{3}+\cdots+x^{27})(1+x+x^{2}+x^{3}+\cdots+x^{14})^{2} \\ = (1+x+x^{2}+\cdots+x^{27}) (1+2x+3x^{2}+\cdots +14x^{13}+15x^{14} +14x^{15} \cdots +2x^{27}+ x^{28}) \\ = \cdots+x^{28}( \underbrace{1+2 +\cdots + 15}_{120} + \underbrace{14+ \cdots +2}_{104})+ \cdots \end{eqnarray*}

EDIT

Answer 3

Express is as a product of three factors $F_1,F_2,F_3$: $$(1+x+x^{2}+x^{3}+....+x^{27})(1+x+x^{2}+x^{3}+....x^{14})^{2}=(1+\cdots +x^{27})(1+\cdots +x^{14})(1+\cdots +x^{14}).$$ Make up a table of powers of $x$: $$\begin{array}{c|c|lcr} m& n & \text{$F_1$} & \text{$F_2$} & \text{$F_3$} \\ \hline & 1 & 14 & 14 & 0 \\ & \downarrow & \downarrow & \downarrow & \vdots \\ 1 & 14 & 27 & 1 & 0 \\ & 1 & 13 & 14 & 1 \\ & \downarrow & \downarrow & \downarrow & \vdots \\ 1 &15 & 27 & 0 & 1 \\ & 1 & 12 & 14 & 2 \\ &\downarrow & \downarrow & \downarrow & \vdots \\ 2 &15 & 26 & 0 & 2 \\ &1 & 11 & 14 & 3 \\ &\downarrow & \downarrow & \downarrow & \vdots \\ 3 & 15 & 25 & 0 & 3 \\ & 1 & 10 & 14 & 4 \\ & \downarrow & \downarrow & \downarrow & \vdots \\ 4 & 15 & 24 & 0 & 4 \\ & 1 & 9 & 14 & 5 \\ & \downarrow & \downarrow & \downarrow & \vdots \\ 5 & 15 & 23 & 0 & 5 \\ & \cdots & \cdots & \cdots & \cdots \\ & 1 & 0 & 14 & 14 \\ & \downarrow & \downarrow & \downarrow & \vdots \\ 14 & 15 & 14 & 0 & 14 \\ \end{array}$$ Hence: $$1\cdot 14 + 14\cdot 15=224.$$

Answer 4

If we multiply out $(1+x+x^2+\cdots+x^{14})^2$ but don't collect like terms yet, there will be $15^2=225$ terms, all with coefficient $1$ (and various exponents between $0$ and $28$).

Each of these $225$ terms can combine with exactly one of the terms of $1+x+x^2+\cdots+x^{27}$ to make $x^{28}$ -- except for $1\cdot 1$ which would need a $x^{28}$ term that isn't in $1+x+\cdots+x^{27}$.

So there are $225-1=224$ terms of $x^{28}$ to add up, giving $224x^{28}$.

Answer 5

The coefficients are bounded by $28 \cdot 15^2 = 6300$. So the following program that uses Python 3.6 integer arithmetic can be used to find the solution

x=(((6301**29-1)*(6301**15-1)**2)//6300**3)
y=x%(6301**28)
z=y//6301**27
print(z)

It prints out

224

Python's operators:

• = is the assignment operator
• * is the multiplication operator
• ** is the exponentiation operator
• % is the remainder operator
• // is the integer division operator