**#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.**

Where to post the proof ? ;)

ReplyDeleteThe 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.

Rito,

ReplyDeletePattern is correct