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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.

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  • $\begingroup$ Hint: $a_n$ is an integer. $\endgroup$
    – Math-fun
    Commented Oct 6, 2018 at 5:31

1 Answer 1

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

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  • $\begingroup$ 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? $\endgroup$
    – Shrey Joshi
    Commented Oct 8, 2018 at 19:20
  • $\begingroup$ Note that: $$\sum_{a_1=1}^9 \sum_{a_2=a_1}^9 \sum_{a_3=a_2}^9 a_3=$$ $\endgroup$
    – farruhota
    Commented Oct 10, 2018 at 14:04
  • $\begingroup$ $$\begin{pmatrix} 1+2+3+\cdots +9 \\ \ \ \ \ \ \ \ 2+3+\cdots+9 \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 3+\cdots+9 \\ \qquad \vdots \\ \qquad \qquad \qquad \ \ \ \ \ 9 \\ \end{pmatrix}+ \begin{pmatrix} 2+3+\cdots +9 \\ \ \ \ \ \ \ \ 3+\cdots+9 \\ \qquad \vdots \\ \qquad \ \ \ \ \ \ \qquad9 \\ \end{pmatrix}+ \begin{pmatrix} 3+\cdots +9 \\ \ \vdots \\ \qquad \qquad9 \\ \end{pmatrix}+\cdots +(9)=\\$$ $\endgroup$
    – farruhota
    Commented Oct 10, 2018 at 14:04
  • $\begingroup$ $$\color{red}{1}\cdot \color{blue}1+\color{red}2\cdot \color{blue}{(2+1)}+\color{red}3\cdot \color{blue}{(3+2+1)}+\cdots +\color{red}9\cdot \color{blue}{(9+8+\cdots +1)}=\\ \sum_{k=1}^9 \color{red}k\cdot \color{blue}{\frac{k(k+1)}{2}}=1155.$$ $\endgroup$
    – farruhota
    Commented Oct 10, 2018 at 14:04
  • $\begingroup$ See also Figurate numbers $\endgroup$
    – farruhota
    Commented Oct 10, 2018 at 14:06

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