Problem Solving

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]

2+2= 2*(2)=2^2
2^2+2^2= 2*2^2=2^3
and so on..
you will finally get 2^9

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.

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

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 ?

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!

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

oops my bad...sorry