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(i) By using sum of terms in a geometric progression formula, the following can be proved: $$ \sum_{a+b=n} x^ay^b = \frac{x^{n+1}-y^{n+1}}{x-y} $$

(ii) Is there a formula for $$ \sum_{a+b+c=n} x^ay^bz^c $$

(iii) If yes, can it be expanded to a general case $$ 

 \sum_{k_{1}+k_{2}+...+k_{m}=n} x_{1}^{k_{1}}x_{2}^{k_{2}}...x_{m}^{k_{m}} $$ My attempt:

I tried treating the last two variables ($y$ and $z$ with their exponents) as a single term (in eqn. (ii)) and applying the first formula (eqn. (i)). However, I couldn't figure out how to deal with the first term ($x^a$), which is varying.

Sebastiano
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Jothin kumar
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2 Answers2

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If we consider the power series in variables, $x,y,z,t$, then $\sum_{a+b+c=n}x^ay^bz^c$ will be the coefficient of $t^n$ in $$(1+xt+x^2t^2+\cdots)(1+yt+y^2t^2+\cdots)(1+zt+z^2t^2+\cdots)$$ $$=\frac{1}{(1-xt)(1-yt)(1-zt)}$$ Your next step is to do partial fraction decomposition. $$\frac{1}{(1-xt)(1-yt)(1-zt)}=\frac{A}{1-xt}+\frac{B}{1-yt}+\frac{C}{1-zt}$$ $$1=A(1-yt)(1-zt)+B(1-xt)(1-zt)+C(1-yt)(1-zt)$$ We can substitute $t=\frac{1}{x},\frac{1}{y},\frac{1}{z}$ to get $$A=\frac{x^2}{(x-y)(x-z)}$$ $$B=\frac{y^2}{(y-x)(y-z)}$$ $$C=\frac{z^2}{(z-x)(z-y)}$$ So the coefficient of $t^n$ will be $$\frac{x^{n+2}}{(x-y)(x-z)}+\frac{y^{n+2}}{(y-x)(y-z)}+\frac{z^{n+2}}{(z-x)(z-y)}$$ You can do the same process for more variables, or simply extrapolate what the expression should be.

Alan Abraham
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Your attempt sounds to be exactly what you should do. First sum for a fixed power of $z$:-

$$\displaystyle\sum\limits_{a+b+c=n} x^ay^bz^c=\sum\limits_{u+c=n} \frac{x^{u+1}-y^{u+1}}{x-y}z^c$$

Then sum for the $x^{u+1}$ and $y^{u+1}$ terms separately:- $$x\displaystyle\sum\limits_{u+c=n} \frac{x^{u}z^c}{x-y}=\frac{x(x^{n+1}-z^{n+1})}{(x-y)(x-z)}$$

Similarly $$y\displaystyle\sum\limits_{u+c=n} \frac{y^{u}z^c}{x-y}=\frac{y(y^{n+1}-z^{n+1})}{(x-y)(y-z)}$$ Combining then gives the sum as $$\frac{x^{n+2}}{(x-y)(x-z)}+\frac{y^{n+2}}{(y-x)(y-z)}+\frac{z^{n+2}}{(z-y)(z-x)}$$

user1172706
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