Number of terms in the expansion $(1+a^3+a^{-3})^{100}$

Question

Find the number of terms in the expansion $(1+a^3+a^{-3})^{100}$

I used the concept $a^3+a^{-3}=T$, while using this I have 101 terms, from $T^2$ to $T^{100}$ how do i find the number of terms that do not intersect

Answer 1

Write $a^3=b$

Multinomial theorem says

The general term of $$(1+b+b^{-1})^{100}$$ is $$\dfrac{100!}{p!q!r!}b^{p-q}$$ with

$p+q+r=100$ and $p,q,r\ge0$

$\implies0\le p,q\le100$

So, $-100\le p-q\le100$

If $d=p-q,100=d+2q+r\iff d=100-2q-r$

If $r=0,p+q=100$

$d=100-2q,d$ can attain all the even integer values in $[-100,100]$

If $r=1,p+q=99$

$d=100-(2q+1)=99-2q$ can attain all the odd values in $[-99,+99]$

Clearly $p-q$ can attain all the possible $100-(-100)+1$ values

Answer 2

Well, note that

$$(a^{-3} + 1+a^3)^{100} = (a^{-3}+a^0+a^{3})^{100} = \sum_{i=-100}^{100} c_ia^{3i},$$

for some $c_i$ and each $c_i$ is strictly positive. In fact, each $c_i$ is precisely the number of ordered multisets $\{b_1,\ldots, b_{100}\}$; $b_l \in \{-1,0,1\}$ such that $i=\sum_{l=1}^{100} b_l$. [Make sure you see why]. This is at least 1 for each $i \in \{-100, -99,\ldots, -1, 0,1, \ldots, 100\}$; indeed if $i$ is positive let $b_l=1$ for each $l=1,2,\ldots, i$ and $b_l=0$ for each other $i$, and if $i$ is negative let $b_l=-1$ for each $i=1,2,\ldots, |i|$, and $b_l=0$ for each other $l$. If $i$ is 0 then let all the $b_l$s are 0.

So, as each $c_i$ is positive this would be 201 terms.

Now with that said $(1+a^3)^{100}$ would have 101 terms. Do you see why?

Answer 3

It is equivalent to finding the number of terms in: $$(1+a^3+a^{-3})^{100}=a^{-300}(a^6+a^3+1)^{100} \Rightarrow (a^6+a^3+1)^{100},$$ whose terms will have powers (since the signs are positive and exponents are multiples of $3$): $$0,3,6,...,597,600$$ The number of terms of the arithmetic progression is: $$n=\frac{a_n-a_1}{d}+1=\frac{600-0}{3}+1=201.$$