Trinomial expansion with power constraints

Question

For the trinomial expansion $(a+b+c)^n$, I'd like to sum up the terms like $a^i b^j c^k$ with the constraint $i>j$. How to calculate it efficiently?

Answer 1

I'll do a small example:

$(a+b+c)^3=((a+b)+c)^3$

by associative property of addition. expanding this gives:

$$(a+b)^3+3(a+b)^2c+3(a+b)c^2+c^3$$

extracting from the binomial expansions $\lfloor\frac{k+1}{2}\rfloor$ terms that have $i>j$ in $(a+b)^k$, gives: $$a^3+3a^2b+3a^2c+3ac^2$$ which is $$(a+c)^3-c^3+3a^2b$$

Multinomial theorem, simply gives a way to calculate the coefficients/full expansion quickly.

EDITS done to expand last paragraph, and fix a few stupid math errors.

Answer 2

Use first the binomial power expansion

\begin{align} (\hat{a}+c)^n=\sum_{k=0}^n{{n}\choose k}\hat{a}^kc^{n-k}. \end{align}

Now substitute $\hat{a}\equiv a+b$ to obtain

\begin{align} (\hat{a}+c)^n&=\sum_{k=0}^n{{n}\choose k}\hat{a}^kc^{n-k}\\ &=\sum_{k=0}^n{{n}\choose k}(a+b)^kc^{n-k} \end{align}

Now, apply the binomial expansion a second time to$(a+b)^k$, ie $(a+b)^k=\sum_{l=0}^k{{k}\choose l}a^lb^{k-l}$,

\begin{align} (\hat{a}+c)^n&=\sum_{k=0}^n{{n}\choose k}\hat{a}^kc^{n-k}\\ &=\sum_{k=0}^n{{n}\choose k}(a+b)^kc^{n-k}\\ &=\sum_{k=0}^n{{n}\choose k}\biggl(\sum_{l=0}^k{{k}\choose l}a^lb^{k-l}\biggr)c^{n-k}\\ &=\sum_{k=0}^n\sum_{l=0}^k{{n}\choose k}{{k}\choose l}a^lb^{k-l}c^{n-k} \end{align}

Your above condition to only sum up terms with $i>j$ translates in my notation to $l>k-l$ or equivalently $l>\tfrac{k}{2}$, thus your power expansion ultimately reduces to

\begin{align} \sum_{k=0}^n\sum_{l>\frac{k}{2}}^k{{n}\choose k}{{k}\choose l}a^lb^{k-l}c^{n-k} \end{align}

Answer 3

Two variations.

We have \begin{align*} (a+b+c)^n&=\sum_{k=0}^n\binom{n}{k}(a+b)^kc^{n-k}\\ &=\sum_{k=0}^n\binom{n}{k}\sum_{j=0}^k\binom{k}{j}a^jb^{k-j}c^{n-k} \end{align*}

It follows \begin{align*} \color{blue}{\sum_{k=0}^n\ \sum_{j=\left\lfloor \frac{k}{2}\right\rfloor+1}^k \binom{n}{k} \binom{k}{j}a^jb^{k-j}c^{n-k}} \end{align*}

We have \begin{align*} (a+b+c)^n&=\sum_{{i+j+k=n}\atop{i,j,k\geq 0}}\binom{n}{i,j,k}a^ib^jc^k \end{align*}

It follows \begin{align*} \color{blue}{\sum_{{i+j+k=n}\atop{i>j,k\geq 0}}\binom{n}{i,j,k}a^ib^jc^k} \end{align*}