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Ask iit jee aieee pet cbse icse state board community Community Discussion Question: mast question on sereis....?????
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[Post New]23 Dec 2007 12:34:51 IST
    Subject: mast question on sereis....?????
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tarun_bits (639)

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hi here is a nice question...
prove that the numbers
49,4489,444889......
obtained by inserting 48 into the middle of the preceding number are squares of integers...
 
i will post the solution...
make ur try geniuses..
<TABLE CELLSPACING="1" CELLPADDING="1" BORDER="0">
<TR><TD>


<DIV ALIGN="right">Glitter Graphics</DIV></TD></TR></TABLE>





    
[Post New]23 Dec 2007 13:09:23 IST
    Subject: mast question on sereis....?????
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tarun_bits (639)

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try it...
<TABLE CELLSPACING="1" CELLPADDING="1" BORDER="0">
<TR><TD>


<DIV ALIGN="right">Glitter Graphics</DIV></TD></TR></TABLE>





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[Post New]23 Dec 2007 13:24:24 IST
    Subject: Re:mast question on sereis....?????
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nadeemoidu (1184)

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Each term can be simplified as [ 2/3(10n) + 1/3 ]2

or [( 2x10n +1)/3]2  .

2x10n +1 is divisible by 3 as the sum of the digits is 3. Hence the answer
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[Post New]23 Dec 2007 13:26:25 IST
    Subject: mast question on sereis....?????
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tarun_bits (639)

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show how..
<TABLE CELLSPACING="1" CELLPADDING="1" BORDER="0">
<TR><TD>


<DIV ALIGN="right">Glitter Graphics</DIV></TD></TR></TABLE>





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[Post New]23 Dec 2007 13:30:07 IST
    Subject: Re:mast question on sereis....?????
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nadeemoidu (1184)

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444....488......89  = 4 ( 111...1)x10n + 8(1111....1) +1
<--n---><--n-1->

1111....1 can be expressed as (10n -1)/9 [ It is the sum of a GP]

Now just simplify
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[Post New]23 Dec 2007 16:49:52 IST
    Subject: mast question on sereis....?????
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tarun_bits (639)

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nicy man......
<TABLE CELLSPACING="1" CELLPADDING="1" BORDER="0">
<TR><TD>


<DIV ALIGN="right">Glitter Graphics</DIV></TD></TR></TABLE>





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