Friday, January 15, 2010

Again a perfect square

#Problem11
49=7^2
4489=67^2
444889=667^2
............................
Now prove that if we insert 48 in the middle of the previous term of this series, like this way, in each case it will produce a perfect square.

Solution scheme and approach 
Number is in this format
[4444...4](n+1 numbers)[8888...8](n numbers)9
So we can write the number in the following way
[4*10^(2n+1)+4*10^2n+4*10^(2n-1)+....4*10^(n+1)] + [8*10^n+8*10^(n-1)+8*10^(n-2)+...8*10]+9
=4*10^(n+1)*[10^(n+1)-1]/9+8*10*(10^n-1)/9+9
=1/9[4*10^(n+1)*{10^(n+1)-1}+80*(10^n-1)+81]
=1/9[{2*10^(n+1)}^2-4*10^(n+1)+8*10^(n+1)-80+81]
=1/9[{2*10^(n+1)}^2-4*10^(n+1)+1]
=1/9[2*10^(n+1)-1]^2
=>a perfect square for n>=0 Q.E.D.

 

2 comments:

  1. Where to post the proof ? ;)

    The pattern is like this : (posted this on Pg also)

    4489 - 49 = 4440 = 60^2 + 2*60*7
    So 4489 = 60^2 + 2*60*7 + 7^2 = (60+7)^2
    Similarly, 444889 - 4489 = 440400 = 600^2 + 2* 600*67
    So 444889 = 600^2 + 2* 600*67 + 67^2 = 667^2

    This is the pattern I guess.....not sure if it's the proof though.

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