To find number of distinct terms of binomial expansion

Question

I have binomial expansion as

$(x+\frac{1}{x} + x^2 + \frac{1}{x^2})^{15}$.

How do i find number of distinct terms in it. Distinct in sense means terms having different powers of $x$?

I have simplified this as $\frac{1}{x^{30}}(x^3+x+x^4+1)^{15}$. How do i proceed

Thanks

Answer 1

$$(x^4+x^3+x+1)^n$$ $$\quad{\text{distinct terms}}=4n+1 \quad{\text{where n>1}}$$ so $$\quad{\text{distinct terms}}=4(15)+1=61$$

Answer 2

$$\frac{1}{x^{30}}(x^3+x+x^4+1)^{15}=\frac{1}{x^{30}}(x^3(x+1)+x+1)^{15}=$$ $$=\frac{1}{x^{30}}(x^3+1)^{15}(x+1)^{15}=\frac{1}{x^{30}}\sum_{i=0}^{15}\binom{15}{i}x^{3i}\sum_{j=0}^{15}\binom{15}{j}x^{j}=$$ $$=\sum_{i=0}^{15}\binom{15}{i}\sum_{j=0}^{15}\binom{15}{j}x^{3i+j-30}$$ Number of elements in set $$\{3i+j-30:i,j\in\{0,1,...,15\}\}=\{-30,-29,...,0,1,...,30\}$$ show the distinct powers of $x$

Answer 3

We have that $$(x^4+x^3+x+1)^n=(x+1)^n\cdot(x^3+1)^n.$$ Now the powers of $x$ in $(1+x)^n$ are the integers in the interval $[0,n]$. Moreover for $n\geq 2$, the multiplication of $(x^3+1)(x+1)^n$ gives a sum of powers of $x$ which cover the range $[0,n]\cup [3,n+3]=[0,n+3]$.

Hence, for $(x+1)^n\cdot(x^3+1)^n$ the powers of $x$ cover the range $[0,n+3n]=[0,4n]$, which contains $4n+1$ integers.

Therefore $$\left(x+\frac{1}{x} + x^2 + \frac{1}{x^2}\right)^{15}=\frac{1}{x^{30}}(x^3+x+x^4+1)^{15}$$ contains all the powers of $x$ in the range $[-30,4\cdot 15-30]=[-30,30]$. Their number is $61$.

Answer 4

Here is a slightly different elementary approach. Expanding the expression \begin{align*} \left(x^{-2}+x^{-1}+x^1+x^2\right)^{15}\tag{1} \end{align*} and keeping the focus on the powers of $x$ means to select $15$ times an element from the set

\begin{align*} A_1:=\{-2,-1,1,2\} \end{align*} and adding them.

We claim the following: There are $61$ distinct terms in the expansion of (1), meaning that in the representation \begin{align*} \left(x^{-2}+x^{-1}+x^1+x^2\right)^{15}=\sum_{k=-30}^{30}a_k x^k \end{align*} all $a_k\ne 0$, $-30\leq k\leq 30$.

We will show by repeatedly adding the set $A_1$ that we get all the integer powers in between and including $-30$ and $30$. In order to do so we define for sets $A,B\subset \mathbb{Z}$, the sum $A+B$ as \begin{align*} A+B:=\{a+b: a\in A, b\in B\} \end{align*}

First step: $A_1+A_1$

At first we calculate $A_2:= A_1+A_1$ which gives all distinct terms of $\left(x^{-2}+x^{-1}+x^1+x^2\right)^2$.

We obtain with $A_1=\{-2,-1,1,2\}$ \begin{align*} A_1+A_1&=\left(\{-2\}+A_1\right)\cup\left(\{-1\}+A_1\right)\cup\left(\{1\}+A_1\right)\cup\left(\{2\}+A_1\right)\\ &=\{-4,-3,-1,0\}\cup\{-3,-2,0,1\}\cup\{-1,0,2,3\}\cup\{0,1,3,4\}\\ &=[-4,4]\cap\mathbb{Z} \end{align*}

Note that $A_2$ contains all integers in $[-4,4]$. Next we show that when we add $A_1$ to such a contiguous set of integers, we increase the set by preserving this kind of contiguity.

Second step: $([-n,n]\cap\mathbb{Z})+A_1=[-n-2,n+2]\cap\mathbb{Z}$

We obtain with $n\geq 1$

\begin{align*} ([-n,n]\cap\mathbb{Z})+A_1&=([-n,n]\cap\mathbb{Z})+\{-2,-1,1,2\}\\ &=\left(([-n,n]\cap\mathbb{Z})+\{-2\}\right)\cup\left(([-n,n]\cap\mathbb{Z})+\{-1\}\right)\\ &\qquad\cup\left(([-n,n]\cap\mathbb{Z})+\{1\}\right)\cup\left(([-n,n]\cap\mathbb{Z})+\{2\}\right)\\ &=\left([-n-2,n-2]\cap\mathbb{Z}\right)\cup\left([-n-1,n-1]\cap\mathbb{Z}\right)\\ &\qquad\cup\left([-n+1,n+1]\cap\mathbb{Z}\right)\cup\left([-n+2,n+2]\cap\mathbb{Z}\right)\\ &=[-n-2,n+2]\cap\mathbb{Z} \end{align*}

We observe when adding $A_1$ to a contiguous set $[-n,n]\cap\mathbb{Z}$ of integers it increases the set by $\{-n-2,-n-1,n+1,n+2\}$ without removing already existing integers.

Final step:

In order to determine the distinct powers of (1) we have to calculate $A_{15}$ which is according to the steps above \begin{align*} A_{15}:=\sum_{j=1}^{15}A_1=[-30,30]\cap\mathbb{Z} \end{align*}

We conclude: The number of distinct terms in (1) is $$\left|A_{15}\right|=\left|[-30,30]\cap\mathbb{Z}\right|=61$$

Note: The core of this answer is the second step which can be rewritten as proof by induction.

Answer 5

We have,

$\sf \bigg( \dfrac{x^4+x^3+x+1}{x^2}\bigg)^{15}=\dfrac{a_0+a_1x+a_2x^2+ ... +a_{60}x^{60}}{x^30}$

Hence, the total number of terms is 61.