Re: Re: finding explicit rule for series

```This is easy to prove. Just write

n[i]-n[i-1]   ==  5342543
n[i-1]-n[i-2] == 5342543
......................
n[2]-n[1]    == 5342543
n[1]-n[0]     == 5342543

Of course you can also prove it by induction.

(I failed to notice posting. Maybe someone has already posted the above
proof, in which case I apologise for the duplication).

Andrzej Kozlowski
Chiba, Japan
http://www.akikoz.net/~andrzej/
http://www.mimuw.edu.pl/~akoz/

On 9 Nov 2004, at 15:36, DrBob wrote:

> *This message was transferred with a trial version of CommuniGate(tm)
> Pro*
> It turns out that n[i] == 5342543 i + n[0].
>
> n[i_?Positive]:=5342543+n[i-1]
> n/@Range@5
> 5342543Range@5+n[0]
>
> {5342543+n[0],10685086+n[0],16027629+n[0],21370172+n[0],26712715+n[0]}
>
> {5342543+n[0],10685086+n[0],16027629+n[0],21370172+n[0],26712715+n[0]}
>
> I calculated a few terms both ways and the answers agree.
>
> Bobby
>
> On Mon, 8 Nov 2004 03:13:28 -0500 (EST), Uwe Ziegenhagen
> <newsgroup at ziegenhagen.info> wrote:
>
>> Hello,
>>
>> i have a series of numbers and would like to use Mathematica to find
>> the
>> explicit algorithm in that way:
>>
>> n(i)=5342543+n(i-1)....
>>
>> I already spent some time with the manual, but didn't find anything.
>>
>> Uwe
>>
>
>
>
> --
> DrBob at bigfoot.com
> www.eclecticdreams.net
>
>

```

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