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IIT-JEE 2009 practise problem (1)

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Original Question by Aeroish

Here's the question...

Prove That the numbers 49, 4489, 444889, .... obtained by insering 48 into the middle of the preceding number are squares of integers.

Posted by Aeroish 09:37, 10 September 2009 (UTC)

Post a reply![1]


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Reply to the thread, posted by Aeroish

Thread post

Additional notes

hint :concepts used while solving this question 

     1.) generalize the terms , say nth term
     2.) use the formulaes for "finding the sum"
These comments, along with the notes on the left, were contributed by Aeroish 09:42, 10 September 2009 (UTC)

Ans

 
Reply to the thread, posted by Pravinsanap

Thread post

Our numbers are: 49 4489 444889 44448889 Hence our general number is: 444444444……..88888888888…....9 (n times) (n-1 times) (1 time) Let it be denoted by S S = 444444444……..88888888888…....9 2n digits 

     (n times)	(n-1 times)	(1 time)

S = 444444444….4*10^n +88888888….8*10 + 9 S = 4*(10^n)*(111111…1 (n digits)) + 8*10(111111…1 (n-1 digits)) + 9 S = 4*(10^n)*(1 + 10 + 10^2 + 10^3 + ………10^n-1) + 8*10(1 + 10 + 10^2 + 10^3 + ………10^n-2) + 9 S = 4(10^n) ((10^n)-1)/ (10-1) + 8(10)((10^n-1)-1)/(10-1) + 9 [Using 1 + 10 + 10^2 + 10^3 + ………10^n-1 = (10^n)-1)/(10-1)] S = (4/9)(10^2n +10^2 + 1/4) [By simplifying] S= [(2/3)(10^2 +1/2)^2] Hence it is proved that S is a perfect square Now we have to prove that it is a perfect square of integers Hence we have to prove that (2/3)(10^n + ½) are integers when n is an integer Put n =1,2,3,4 and notice that (2/3)(10^n + ½) are integers

Additional notes

These comments, along with the notes on the left, were contributed by Pravinsanap 13:03, 12 April 2010 (UTC)
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