- #1

yellowsnow

- 4

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## Homework Statement

For all

*n*∈

**Z**

^{+}, the function

*P*of

_{n}*i*variables is defined recursively as follows:

*P*(

_{n}*x*

_{1},...,

*x*

_{n}) =

*P*

_{n}_{-1}(

*x*

_{1}+

*x*

_{2},

*x*

_{2}+

*x*

_{3},...,

*x*

_{n}_{-1}+

*x*) and

_{n}*P*1(

*x*

_{1}) =

*x*

_{1}.

Find a closed formula for

*P*.

_{n}## Homework Equations

*P*(

_{n}*x*

_{1},...,

*x*

_{n}) =

*P*

_{n}_{-1}(

*x*

_{1}+

*x*

_{2},

*x*

_{2}+

*x*

_{3},...,

*x*

_{n}_{-1}+

*x*) and

_{n}*P*1(

*x*

_{1}) =

*x*

_{1}.

## The Attempt at a Solution

So far, I've found it definitely follows a Pascal's Triangle-esque pattern:

*P*

_{1}(

*x*

_{1}) =

*x*

_{1}

*P*

_{2}(

*x*

_{1},

*x*

_{2}) =

*P*

_{1}(

*x*

_{1}+

*x*

_{2}) =

*x*

_{1}+

*x*

_{2}

*P*

_{3}(

*x*

_{1},

*x*

_{2},

*x*

_{3}) =

*P*

_{2}(

*x*

_{1}+

*x*

_{2},

*x*

_{2}+

*x*

_{3}) =

*P*

_{1}(

*x*

_{1}+ 2

*x*

_{2}+

*x*

_{3}) =

*x*

_{1}+ 2

*x*

_{2}+

*x*

_{3}

*P*

_{4}(

*x*

_{1},

*x*

_{2},

*x*

_{3},

*x*

_{4}) =

*P*

_{3}(

*x*

_{1}+

*x*

_{2},

*x*

_{2}+

*x*

_{3},

*x*

_{3}+

*x*

_{4}) =

*P*

_{2}(

*x*

_{1}+ 2

*x*

_{2}+

*x*

_{3},

*x*

_{2}+ 2

*x*

_{3}+

*x*

_{4}) =

*P*

_{1}(

*x*

_{1}+ 3

*x*

_{2}+ 3

*x*

_{3}+

*x*

_{4}) =

*x*

_{1}+ 3

*x*

_{2}+ 3

*x*

_{3}+

*x*

_{4}

I know it's very similar to Pascal's triangle, but I'm just not sure how to find a closed form for it.

I'd appreciate any help; thanks!