SOLUTION: The least value of n for which sum of n terms of series 1+3+9(3square)+.....is greater than 7000 is

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Question 878997: The least value of n for which sum of n terms of series 1+3+9(3square)+.....is greater than 7000 is

Answer by AnlytcPhil(1807)   (Show Source): You can put this solution on YOUR website!
It's a geometric series with a1 = 1 and r = 3.

The sum of a geometric series is given by the formula:



We set that greater than 7000







Multiply both sides by 2 to clear the fraction:



Add 1 to both sides



Take the ln or log of both sides:



Use the property of logs that says that the log of 
an exponential is the product of the exponent and 
the log of the base.



Divide both sides by log(3)





The least value of n greater than that is when n=9.

Edwin
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