Number of terms in the expansion $(1+a^3+a^{-3})^{100}$ 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
 A: 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
A: 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?
A: 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:
$$n=\frac{a_n-a_1}{d}+1=\frac{600-0}{3}+1=201.$$
