I want to figure out this series and represent it as a nested sum

Question

I want to figure out the series on k, where k is a positive integer. is it possible to make a single definition for every k, so we just have to substitute the value of k to get the series of each k.

$k=0$ ;

$\;\;ABC$

$k=1$ ;

$\;\;A^{4}BC+A^{3}B^{2}C+A^{3}B^{}C^{2}+A^{2}B^{3}C^{}+A^{2}B^{2}C^{2}$

$k=2$ ;

$\;\;A^{7}BC+A^{6}B^{2}C+A^{6}B^{}C^{2}+A^{5}B^{3}C^{}+A^{5}B^{2}C^{2}+A^{5}B^{}C^{3}+A^{4}B^{4}C^{}+A^{4}B^{3}C^{2}+A^{4}B^{2}C^{3}+A^{3}B^{5}C^{}+A^{3}B^{4}C^{2}+A^{3}B^{3}C^{3}$

$k=3$ ;

$\;\;A^{10}BC+A^{9}B^{2}C+A^{9}B^{}C^{2}+A^{8}B^{3}C^{}+A^{8}B^{2}C^{2}+A^{8}B^{}C^{3}+A^{7}B^{4}C^{}+A^{7}B^{3}C^{2}+A^{7}B^{2}C^{3}+A^{7}B^{}C^{4}+A^{6}B^{5}C^{}+A^{6}B^{4}C^{2}+A^{6}B^{3}C^{3}+A^{6}B^{2}C^{4}+A^{5}B^{6}C^{}+A^{5}B^{5}C^{2}+A^{5}B^{4}C^{3}+A^{5}B^{3}C^{4}+A^{4}B^{7}C^{}+A^{4}B^{6}C^{2}+A^{4}B^{5}C^{3}+A^{4}B^{4}C^{4}$

Thanks in advance for any help!

This is what I have got

$$\sum_{j=1}^{k} \sum_{i=1}^{j} A^{3k+2-j} B^{1-i+j} C^i + \sum_{i=1}^{k+1} \sum_{j=1}^{k+1} A^{2k+2-j} B^{k+1-i+j} C^i$$

But I want it as a single nested sum.

Answer 1

Observations

• $n=1$ ; $A^1B^1C^1 \to 1+1+1=3$
• $n=2$ ; $A^{4}B^1C^1+A^{3}B^{2}C+A^{3}B^{}C^{2}+A^{2}B^{3}C^{}+A^{2}B^{2}C^{2}\to 4+1+1=3+2+1=2+2+2=3\times2$
• $\ldots$

Seems like for $n$, we have positive integer partitions of $3n$. Though power of $A$ doesn't fall more than $n$.

Hence,

$$T_n = \sum_{(i,j,k) \in S} A^i B^j C^k $$

where $$S:= \{(i,j,k)| i+j+k=3n ; \, i\ge n ; \, i,j,k \in \mathbb{N} \}$$