Addin Exponents

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Addin Exponents

by heshamelaziry » Thu Aug 27, 2009 5:44 pm
Could you Help with this one:

2+2+2^2+2^3+2^4+2^5+2^6+2^7+2^8


Answer is 2^9

Could you explain ???????[/b][/u][/quote][/list][/list]

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by tohellandback » Thu Aug 27, 2009 9:06 pm
2+2= 2*(2)=2^2
2^2+2^2= 2*2^2=2^3
and so on..
you will finally get 2^9
The powers of two are bloody impolite!!

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by heshamelaziry » Thu Aug 27, 2009 10:09 pm
tohellandback wrote:2+2= 2*(2)=2^2
2^2+2^2= 2*2^2=2^3
and so on..
you will finally get 2^9

Could you explain this further? Thanks.

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by tohellandback » Thu Aug 27, 2009 10:24 pm
heshamelaziry wrote:
tohellandback wrote:2+2= 2*(2)=2^2
2^2+2^2= 2*2^2=2^3
and so on..
you will finally get 2^9

Could you explain this further? Thanks.
2*(2)=2^2
2^2+2^2= 2*2^2=2^3
2^3+2^3=2*2^3=2^4
2^4+2^4=2*2^4=2^5
2^5+2^5=2*2^5=2^6
2^6+2^6=2*2^6=2^7
2^7+2^7=2*2^7=2^8
2^8+2^8=2*2^8=2^9
The powers of two are bloody impolite!!

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by heshamelaziry » Thu Aug 27, 2009 10:53 pm
tohellandback wrote:
heshamelaziry wrote:
tohellandback wrote:2+2= 2*(2)=2^2
2^2+2^2= 2*2^2=2^3
and so on..
you will finally get 2^9

Could you explain this further? Thanks.
2*(2)=2^2
2^2+2^2= 2*2^2=2^3
2^3+2^3=2*2^3=2^4
2^4+2^4=2*2^4=2^5
2^5+2^5=2*2^5=2^6
2^6+2^6=2*2^6=2^7
2^7+2^7=2*2^7=2^8
2^8+2^8=2*2^8=2^9
Thank you so much. It seems likely that there is a formula for problems like this. Is there one ?

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formula for this

by anand0408 » Sat Aug 29, 2009 7:08 am
if u observe, its actually a GP (geometric Progression) except the first term

therefore: 2 + (2 + 2^2 + 2^3....+ 2^8)

Sum to n terms of a GP is: a[(r^n) - 1] where: a = first term; r = common factor

here: a = 2
r = 2
n=8

therefore: Sum to 8 terms is: 2[(2^8)-1]

Plus we have to add the extra 2 in the beginning:

therefore answer is:

2+ 2[(2^8)-1]
=2[(2^80-1+1]
=2(2^8)
=2^9

njoy!

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Re: formula for this

by rish » Sat Aug 29, 2009 8:33 am
anand0408 wrote:if u observe, its actually a GP (geometric Progression) except the first term

therefore: 2 + (2 + 2^2 + 2^3....+ 2^8)

Sum to n terms of a GP is: a[(r^n) - 1] where: a = first term; r = common factor

here: a = 2
r = 2
n=8

therefore: Sum to 8 terms is: 2[(2^8)-1]

Plus we have to add the extra 2 in the beginning:

therefore answer is:

2+ 2[(2^8)-1]
=2[(2^80-1+1]
=2(2^8)
=2^9

njoy!
Please note

Sum of n terms of a GP is: a[(r^n) - 1]/(r - 1).

Since r is 2 in this case, it doesnt make a difference

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by anand0408 » Sat Aug 29, 2009 12:39 pm
oops my bad...sorry