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Jinglander
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Exponenent Simplification
Source: Beat The GMAT — Problem Solving |
2+(2+...)Jinglander wrote:can someone show me how this was done
2+2+2^2+2^3+2^4+2^5+2^6+2^7+2^8=2^9
This is a GP with 2 as initial term and 2 as ratio.
Sum = a(r^n -1)/r-1
Sum = 2(2^8-1)/2-1
= 2^9 -2
Hence total - 2+2^9-2 = 2^9
HTH
"Once you start working on something,
don't be afraid of failure and don't abandon it.
People who work sincerely are the happiest."
Chanakya quotes (Indian politician, strategist and writer, 350 BC-275BC)
don't be afraid of failure and don't abandon it.
People who work sincerely are the happiest."
Chanakya quotes (Indian politician, strategist and writer, 350 BC-275BC)
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Jinglander
- Senior | Next Rank: 100 Posts
- Posts: 58
- Thanked: 1 times
I dont know if GP is tested on GMAT. But, here is the info.Jinglander wrote:Could really use some more info here on GPs
https://www.math10.com/en/algebra/geomet ... ssion.html
Check this link and post back if you have any queries.
Cheers!!
"Once you start working on something,
don't be afraid of failure and don't abandon it.
People who work sincerely are the happiest."
Chanakya quotes (Indian politician, strategist and writer, 350 BC-275BC)
don't be afraid of failure and don't abandon it.
People who work sincerely are the happiest."
Chanakya quotes (Indian politician, strategist and writer, 350 BC-275BC)
Hello,
I do not see that it requires knowledge of GPs to solve this problem as GPs are not included in the testing material of what i know!
This is how i solved this:
2+2+2^2+2^3+2^4+2^5+2^6+2^7+2^8=
2(1+1+2+2^2+2^3+2^4+2^5+2^6+2^7)=
2(4+2^2+2^3+2^4+2^5+2^6+2^7)=
2(2(2+2+2^2+2^3+2^4+2^5+2^6))=
2(2(4+2^2+2^3+2^4+2^5+2^6))=
2(2(2(2+2+2^2+2^3+2^4+2^5)))=
2(2(2(4+2^2+2^3+2^4+2^5)))=
2(2(2(2(2+2+2^2+2^3+2^4))))=
2(2(2(2(4+2^2+2^3+2^4))))=
2(2(2(2(2(2+2+2^2+2^3)))))=
2(2(2(2(2(4+2^2+2^3)))))=
2(2(2(2(2(2(2+2+2^2))))))=
2(2(2(2(2(2(4+2^2))))))=
2(2(2(2(2(2(2(2+2)))))))=
2(2(2(2(2(2(2(4)))))))=2^9
it might take the whole 2 minute to do this, but this is incase you encounter such a problem with no prior GP knowledge.
Goodluck.
I do not see that it requires knowledge of GPs to solve this problem as GPs are not included in the testing material of what i know!
This is how i solved this:
2+2+2^2+2^3+2^4+2^5+2^6+2^7+2^8=
2(1+1+2+2^2+2^3+2^4+2^5+2^6+2^7)=
2(4+2^2+2^3+2^4+2^5+2^6+2^7)=
2(2(2+2+2^2+2^3+2^4+2^5+2^6))=
2(2(4+2^2+2^3+2^4+2^5+2^6))=
2(2(2(2+2+2^2+2^3+2^4+2^5)))=
2(2(2(4+2^2+2^3+2^4+2^5)))=
2(2(2(2(2+2+2^2+2^3+2^4))))=
2(2(2(2(4+2^2+2^3+2^4))))=
2(2(2(2(2(2+2+2^2+2^3)))))=
2(2(2(2(2(4+2^2+2^3)))))=
2(2(2(2(2(2(2+2+2^2))))))=
2(2(2(2(2(2(4+2^2))))))=
2(2(2(2(2(2(2(2+2)))))))=
2(2(2(2(2(2(2(4)))))))=2^9
it might take the whole 2 minute to do this, but this is incase you encounter such a problem with no prior GP knowledge.
Goodluck.
Best Regards,
Hala Absi
Senior Software Engineer
Hala Absi
Senior Software Engineer
