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by beater » Tue Sep 30, 2008 6:53 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. 102^k − 1

I was able to backsolve and get the answer, but got stuck while simpifying the orginal equation. please help me solve the equation.
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by 4meonly » Tue Sep 30, 2008 7:22 am
I think answer should be A, if I understood Q correctly
(2^k)*(5^(k − 1))?

2*(2^(k-1))*(5^(k − 1))

2*10^(k-1)

OA?
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by beater » Tue Sep 30, 2008 7:47 am
OA - A. Thanks..

Is there another way to do this problem?
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by stop@800 » Tue Sep 30, 2008 9:22 am
beater wrote:OA - A. Thanks..

Is there another way to do this problem?
but the q says (5^k − 1) and not (5^(k − 1))
is this a typo?
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by cramya » Tue Sep 30, 2008 5:12 pm
A. 2(10^k − 1)

Choice A must really be 2 * (10 ^(k-1))

= 2 * ((2*5) ^ (k-1))
= 2 * 2 ^ k-1 * 5 ^ k-1
= 2 ^ k * 5 ^ k-1
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by 4meonly » Wed Oct 01, 2008 7:19 am
stop@800 wrote:
beater wrote:OA - A. Thanks..

Is there another way to do this problem?
but the q says (5^k − 1) and not (5^(k − 1))
is this a typo?
I suppose, yes
I was not able to get any offered answer from (2^k)(5^k − 1)
It should be
(2^k)*(5^(k − 1))?
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by rabab » Wed Oct 01, 2008 9:31 am
i don't get it. can someone clarify pls?
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by rabab » Wed Oct 01, 2008 9:35 am
so okay, back solving from answer A helps. but is there any straight cut way to do it?
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by vishubn » Wed Oct 01, 2008 9:38 am
i don't get it. can someone clarify pls?

as solved earlier ! key point here is to bring

2^k also in the form k-1 so that common factor can be take ouy

so below it has been done just that

2*(2^(k-1))*(5^(k − 1))

try solvoing it u wil get the question

2(2^k/2)*(5^(k − 1))


so going back to equation

2*(2^(k-1))*(5^(k − 1))

k-1 factor out.... powers being same base is multiplied
u get
2*10^k-1 which is one the option



so u get it now???

HTH
vishu[/quote]
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