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

2+(2+...)
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

Could really use some more info here on GPs

Jinglander wrote:Could really use some more info here on GPs

I dont know if GP is tested on GMAT. But, here is the info.

https://www.math10.com/en/algebra/geomet ... ssion.html

Check this link and post back if you have any queries.


Cheers!!

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.