- #1
ThomasMagnus
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A school phone tree has 1 person responsible for contacting 3 people. If there are 1500 students in the school, how many levels will there be on the phone tree (assuming 1 person is at the top of the tree)?
My Solution:
This question forms a geometric series:
A(first term)=1
R(common ratio)=3
1+3+9+27...
Let n= # of levels to the phone tree
When will the sum of the series equal 1500?
Sn=A(1-R^n)/(1-R)
1500=1(1-(3^n))/(1-(3))
-3000=1-(3)^n
-3001=-(3^n)
3001=(3^n)
Log(3001)=nLog(3)
Log(3001)/Log(3)
~7.29
An 8th level would have to be added to the tree; however, the level would not be complete. 8 levels to the tree.
Is this the correct solution to this question? Can you identify where and if I have gone wrong?
Thanks! Appreciate the help :)
My Solution:
This question forms a geometric series:
A(first term)=1
R(common ratio)=3
1+3+9+27...
Let n= # of levels to the phone tree
When will the sum of the series equal 1500?
Sn=A(1-R^n)/(1-R)
1500=1(1-(3^n))/(1-(3))
-3000=1-(3)^n
-3001=-(3^n)
3001=(3^n)
Log(3001)=nLog(3)
Log(3001)/Log(3)
~7.29
An 8th level would have to be added to the tree; however, the level would not be complete. 8 levels to the tree.
Is this the correct solution to this question? Can you identify where and if I have gone wrong?
Thanks! Appreciate the help :)