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Path: utzoo!utcs!lsuc!pesnta!amdcad!decwrl!decvax!dartvax!chuck
From: chuck@dartvax.UUCP (Chuck Simmons)
Newsgroups: net.math
Subject: Re: Ashby's solution to 1+11+...n 1's seems to be in error
Message-ID: <2767@dartvax.UUCP>
Date: Sun, 17-Feb-85 17:11:32 EST
Article-I.D.: dartvax.2767
Posted: Sun Feb 17 17:11:32 1985
Date-Received: Mon, 18-Feb-85 10:16:59 EST
References: <595@decwrl.UUCP>
Organization: Dartmouth College, Hanover, NH
Lines: 29

>> |      Solution to  Sn = 1 + 11 + 111 + ... + 11...11                     |
>> |                                             |<--->|                     |
>> |                                              n 1's                      |
>> |      is easily obtained by re-writing as                                |
>> |                                                                         |
>> |      (1)  Sn = (10^1 - 1)/9 + (10^2 - 1)/9 + ... + (10^n - 1)/9         |
>> |                                                                         |
>> |      re-arranging terms gives                                           |
>> |                                                                         |
>> |      (2)  Sn = 10*(10^n -1)/81  -  n/9                                  |
>> |                                                                         |
>> ===========================================================================
>  
> There's something wrong with this solution !  s4 = 1+11+111+1111 = 1234.
> Let's  check: s4 = 10*(10^4-1)/81 - 4/9 = 10*9999/81 - 4/9 = 1234 + 16/81 -
> 4/9 DOES NOT EQUAL 1234 !!  [(4/9)^2 = 16/81 but so what]

Oops!  10*9999/81-4/9 = (11110-4)/9 = 11106/9 = 1234

Ashby's solution is quite nice.  We have:

Sn = (10^1 - 1)/9 + (10^2 - 1)/9 + ... + (10^n - 1)/9
   = 1/9 * ( (10^1-1) + (10^2-1) + ... + (10^n-1) )
   = 1/9 * (sigma i=1..n of 10^i   - n)
   = 1/9 * (10 * sigma i=0..n-1 of 10^i  - n)
   = 1/9 * (10 * (10^n - 1)/9 - n)
   = 10*(10^n - 1)/ 81 - n/9

as desired.