Let $$f(n)=\sum_{a_1=2}^{9}{\sum_{a_2=a_1}^{9}{\sum_{a_3=a_2}^{9}{...\sum_{a_n=a_{n-1}}^{9}{a_n}}}}$$
A) How could one find $$\sum_{k=1}^{n}{f(k)}?$$ B) How could one find how many terms there are in the sum?
For Part B, I know that the number of terms in $f(n)$ is $f(n-1)$ but I need a way to compute $f(n)$ and the sum in part A.
Note that: $$\sum_{a_1=1}^9 a_1=\sum_{k=1}^9 k=45;\\ \sum_{a_1=1}^9 \sum_{a_2=a_1}^9 a_2=\sum_{k=1}^9 k\cdot k=285;\\ \sum_{a_1=1}^9 \sum_{a_2=a_1}^9 \sum_{a_3=a_2}^9 a_3=\sum_{k=1}^9 k\cdot \frac{k(k+1)}{2}=1155;\\ \sum_{a_1=1}^9 \sum_{a_2=a_1}^9 \sum_{a_3=a_2}^9 \sum_{a_4=a_3}^9 a_4=\sum_{k=1}^9 k\cdot \frac{k(k+1)(k+2)}{6}=3663;\\ \vdots \\ \sum_{a_1=1}^{9}{\sum_{a_2=a_1}^{9}{\sum_{a_3=a_2}^{9}{...\sum_{a_n=a_{n-1}}^{9}{a_n}}}}=\sum_{k=1}^9 k\cdot \frac{k(k+1)(k+2)\cdots(k+n-2)}{(n-1)!}.$$