# 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)
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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) :
* reply

## 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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