Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Maths problem (medium)

  1. May 30, 2010 #1
    mkdagi.jpg

    Use English to explain what this is showing.
    Also prove it!!
     
  2. jcsd
  3. May 30, 2010 #2
    I found this else where, so some of you may have seen it before.
     
  4. May 31, 2010 #3
    Just expand the square and you are basically done. The i's in the rhs are the number of combinations which can produce that exact exponent.
     
  5. May 31, 2010 #4

    HallsofIvy

    User Avatar
    Science Advisor

    The simplest thing to do is to look at what it says for some small n- say n= 3.
    [tex]\left(\sum_{i=0}^{3-1}10^i\right)^2= \left(10^0+ 10^1+ 10^2\right)^2= (1+ 10+ 100)^2= 111^2[/tex].

    [tex]\sum_{i=1}^n i10^{i-1}= 1(10^0)+ 2(10^1)+ 3(10^2)= 321[/tex]

    [tex]\sum_{i=1}^{n-1} i10^{2n-i-1}= 1(10^4}+ 2(10^3)= 1200[/tex]

    It's easy to calculate that [itex]111^2= 12321= 1200+ 321[/itex].

    [itex]11^2= 121[/itex], [itex]111^3= 12321[/itex], [itex]1111^2= 1234321[/itex] , etc.

    Do you see the pattern?
    [tex]\sum_{i=1}^{n-1} i10^{2n-i-1}[/tex]
    is the first part- the 1234... Do you see how it is counting "down" because of the [itex]10^{2n-i-1}[/itex]?
    [tex]\sum_{i=1}^n i10^{i-1}[/tex]
    is the last part: 321
     
  6. May 31, 2010 #5
    Yes thats basically it the first bit gives the last n digits and the first bit gives the first n+1 digits where the lenght is 2n+1
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook