2

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.

Shrey Joshi
  • 596

1 Answers1

1

Note that: $$\sum_{a_1=1}^9 a_1=\sum_{k=1}^9 k=45;\\ \sum_{a_1=1}^9 \sum_{a_2=a_1}^9 a_2=\sum_{k=1}^9 k\cdot k=285;\\ \sum_{a_1=1}^9 \sum_{a_2=a_1}^9 \sum_{a_3=a_2}^9 a_3=\sum_{k=1}^9 k\cdot \frac{k(k+1)}{2}=1155;\\ \sum_{a_1=1}^9 \sum_{a_2=a_1}^9 \sum_{a_3=a_2}^9 \sum_{a_4=a_3}^9 a_4=\sum_{k=1}^9 k\cdot \frac{k(k+1)(k+2)}{6}=3663;\\ \vdots \\ \sum_{a_1=1}^{9}{\sum_{a_2=a_1}^{9}{\sum_{a_3=a_2}^{9}{...\sum_{a_n=a_{n-1}}^{9}{a_n}}}}=\sum_{k=1}^9 k\cdot \frac{k(k+1)(k+2)\cdots(k+n-2)}{(n-1)!}.$$

farruhota
  • 31,482
  • Could someone please help me understand why it is true that $\sum_{a_1=1}^9 \sum_{a_2=a_1}^9 \sum_{a_3=a_2}^9 a_3=\sum_{k=1}^9 k\cdot \frac{k(k+1)}{2}$ and how you can generalize? – Shrey Joshi Oct 08 '18 at 19:20
  • Note that: $$\sum_{a_1=1}^9 \sum_{a_2=a_1}^9 \sum_{a_3=a_2}^9 a_3=$$ – farruhota Oct 10 '18 at 14:04
  • $$\begin{pmatrix} 1+2+3+\cdots +9 \ \ \ \ \ \ \ \ 2+3+\cdots+9 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 3+\cdots+9 \ \qquad \vdots \ \qquad \qquad \qquad \ \ \ \ \ 9 \ \end{pmatrix}+ \begin{pmatrix} 2+3+\cdots +9 \ \ \ \ \ \ \ \ 3+\cdots+9 \ \qquad \vdots \ \qquad \ \ \ \ \ \ \qquad9 \ \end{pmatrix}+ \begin{pmatrix} 3+\cdots +9 \ \ \vdots \ \qquad \qquad9 \ \end{pmatrix}+\cdots +(9)=\$$ – farruhota Oct 10 '18 at 14:04
  • $$\color{red}{1}\cdot \color{blue}1+\color{red}2\cdot \color{blue}{(2+1)}+\color{red}3\cdot \color{blue}{(3+2+1)}+\cdots +\color{red}9\cdot \color{blue}{(9+8+\cdots +1)}=\ \sum_{k=1}^9 \color{red}k\cdot \color{blue}{\frac{k(k+1)}{2}}=1155.$$ – farruhota Oct 10 '18 at 14:04
  • See also Figurate numbers – farruhota Oct 10 '18 at 14:06