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by vongdn » Sat Oct 02, 2010 10:30 pm
2 + 2+ 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8 = ?

A. 2^9
B. 2^10
C. 2^16
D. 2^35
E. 2^37


Outside of calculating this all through, then factoring all the 2's and finding out it is 2 raised to a power, is there a quicker method in finding this?
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by neerajkumar1_1 » Sat Oct 02, 2010 10:43 pm
2 + 2+ 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8

= 2^2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8
= 2^3 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8
and so on so forth...

finally u will get 2^8 + 2^8 = 2^9

IMO A

basically see how I have combined the underlined portions of the equation... and how the sum equals to the next higher power of 2...

I would suggest u revise exponents a bit and then try this problem again..
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by gmatmachoman » Sat Oct 02, 2010 10:45 pm
This is a geometric sequence series:

Sum of the series : 2+ 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8

Sn = A1 ( 1- r^n)/ ( 1-r)

A1 = 2

r = An/ An-1
= 4/2 = 2

Sn = 2 ( 1- 2^ 8)/ ( 1-2)

= 2 ( 2^8-1)

We have excluded the first term 2 for computational purpose

2+ Sn

2 + 2 ( 2^8-1)

2( 1 +2^8 -1)

2^9

Pick A
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by vongdn » Sat Oct 02, 2010 11:13 pm
neerajkumar1_1 wrote:2 + 2+ 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8

= 2^2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8
= 2^3 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8
and so on so forth...

finally u will get 2^8 + 2^8 = 2^9

IMO A

basically see how I have combined the underlined portions of the equation... and how the sum equals to the next higher power of 2...

I would suggest u revise exponents a bit and then try this problem again..
I see how 2^2 + 2^2 = 2^3 and so forth from a numerical aspect, ie if I calculate it out. But that isn't a rule is it? for instance 3^2 + 3^2 does not equal 3^3.
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by neerajkumar1_1 » Sat Oct 02, 2010 11:19 pm
vongdn wrote:
neerajkumar1_1 wrote:2 + 2+ 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8

= 2^2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8
= 2^3 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8
and so on so forth...

finally u will get 2^8 + 2^8 = 2^9

IMO A

basically see how I have combined the underlined portions of the equation... and how the sum equals to the next higher power of 2...

I would suggest u revise exponents a bit and then try this problem again..
I see how 2^2 + 2^2 = 2^3 and so forth from a numerical aspect, ie if I calculate it out. But that isn't a rule is it? for instance 3^2 + 3^2 does not equal 3^3.
oh no no no... see when i calculate 2^2 + 2^2 = 2(2^2) = 2^3

but 3^2 + 3^2 = 2(3^2)... that certainly does not equal to 3^3

but if it was 3^2 + 3^2 + 3^2 = 3.(3^2) = 3^3...

as long as u get the math behind what i have just written... then no problems.. .
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by vongdn » Sat Oct 02, 2010 11:31 pm
neerajkumar1_1 wrote:
vongdn wrote:
neerajkumar1_1 wrote:2 + 2+ 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8

= 2^2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8
= 2^3 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8
and so on so forth...

finally u will get 2^8 + 2^8 = 2^9

IMO A

basically see how I have combined the underlined portions of the equation... and how the sum equals to the next higher power of 2...

I would suggest u revise exponents a bit and then try this problem again..
I see how 2^2 + 2^2 = 2^3 and so forth from a numerical aspect, ie if I calculate it out. But that isn't a rule is it? for instance 3^2 + 3^2 does not equal 3^3.
oh no no no... see when i calculate 2^2 + 2^2 = 2(2^2) = 2^3

but 3^2 + 3^2 = 2(3^2)... that certainly does not equal to 3^3

but if it was 3^2 + 3^2 + 3^2 = 3.(3^2) = 3^3...

as long as u get the math behind what i have just written... then no problems.. .
that makes perfect sense now.....thank you
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