HMMT February

Note that, since this is symmetric in $a_{1}$ through $a_{7}$, $$\sum_{a_{1}=0}^{\infty} \sum_{a_{2}=0}^{\infty} \cdots \sum_{a_{7}=0}^{\infty} \frac{a_{1}+a_{2}+\cdots+a_{7}}{3^{a_{1}+a_{2}+\cdots+a_{7}}} =7 \sum_{a_{1}=0}^{\infty} \sum_{a_{2}=0}^{\infty} \cdots \sum_{a_{7}=0}^{\infty} \frac{a_{1}}{3^{a_{1}+a_{2}+\cdots+a_{7}}} =7\left(\sum_{a_{1}=0}^{\infty} \frac{a_{1}}{3^{a_{1}}}\right)\left(\sum_{a=0}^{\infty} \frac{1}{3^{a}}\right)^{6}$$ If $S=\sum \frac{a}{3^{a}}$, then $3 S-S=\sum \frac{1}{3^{a}}=3 / 2$, so $S=3 / 4$. It follows that the answer equals $7 \cdot \frac{3}{4} \cdot\left(\frac{3}{2}\right)^{6}=\frac{15309}{256}$. Alternatively, let $f(z)=\sum_{a_{1}=0}^{\infty} \sum_{a_{2}=0}^{\infty} \cdots \sum_{a_{7}=0}^{\infty} z^{a_{1}+a_{2}+\cdots+a_{7}}$. Note that we can rewrite $f(z)=\left(\sum_{a=0}^{\infty} z^{a}\right)^{7}=\frac{1}{(1-z)^{7}}$. Furthermore, note that $z f^{\prime}(z)=\sum_{a_{1}=0}^{\infty} \sum_{a_{2}=0}^{\infty} \cdots \sum_{a_{7}=0}^{\infty}\left(a_{1}+a_{2}+\cdots+\right. \left.a_{7}\right) z^{a_{1}+a_{2}+\cdots+a_{7}}$, so the sum in question is simply $\frac{f^{\prime}(1 / 3)}{3}$. Since $f^{\prime}(x)=\frac{7}{(1-z)^{8}}$, it follows that the sum is equal to $\frac{7 \cdot 3^{7}}{2^{8}}=\frac{15309}{256}$.