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(2^k)(5^k-1)=?

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(2^k)(5^k-1)=?

by bhumika.k.shah » Mon Jan 18, 2010 1:43 am
which of the following is equal to (2^k)(5^k-1)?

A.2^(10^k-1)
B.5^(10^k-1)
C.10^k
D.2(10^k)
E.10^2k-1

OA E
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by ksundar » Mon Jan 18, 2010 3:15 am
Are you sure of the option....??

Above equation can be deduced to

=2^K * 5^k/5
= 10^k/5
= 2* 10^k-1

Or am i wrong?
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by bhumika.k.shah » Mon Jan 18, 2010 3:24 am
sowree :(
the OA is A
how did you get the final step ? where did the 2 come from? :-P
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by ramsharma » Mon Jan 18, 2010 9:42 am
bhumika.k.shah wrote:sowree :(
the OA is A
how did you get the final step ? where did the 2 come from? :-P
In the final step he multiplied both numerator and denominator by 2

= 10^k/5
=2*10^k/(2*5)
=2*10^k/10
=2*10^k-1
RAM SHARMA
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by diegow77 » Mon Jan 18, 2010 10:21 pm
ramsharma,

I don't get how you come from 10^k/5 to 2*10^k-1 can you explain in more details.
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by ace_gre » Mon Jan 18, 2010 10:38 pm
Hi, This is my take..

Once we have 10^k/5
==>2^k * 5 ^k / 5 ( can be written as 2^k = 2^k-1 * 2)
==> 2^k-1 * 2 * 5^k-1
==10 ^ k-1 * 2

Hope this helps!
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by diegow77 » Mon Jan 18, 2010 11:19 pm
so you're telling me 5^k/5 is equivalent to 5^k-1?
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by ramsharma » Tue Jan 19, 2010 3:36 am
diegow77 wrote:ramsharma,

I don't get how you come from 10^k/5 to 2*10^k-1 can you explain in more details.

= 10^k/5

=2*10^k/(2*5) -multiplied numerator and denominator by 2(eg 3/5=2*3/5*2)

=2*10^k/10 -Denominator(2*5=10),Numerator remains as it is(eg 3/5=2*3/5*2=2*3/10)

=2*10^k-1 (10^k/10=10^k-1 eg.10^11/10=10^(11-1)

Hope you will get now
RAM SHARMA
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by diegow77 » Wed Jan 20, 2010 12:07 am
Thank you very much, I understand now :)
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