SOLUTION: Evaluate the series if it converges. 9 + 27/4 + 81/16 + 243/64

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Question 39436: Evaluate the series if it converges.
9 + 27/4 + 81/16 + 243/64
Found 2 solutions by AnlytcPhil, astromathman:
Answer by AnlytcPhil(1806)   (Show Source): You can put this solution on YOUR website!
9 + 27/4 + 81/16 + 243/64

Find r by dividing every given term by the term before it

27/4 divided by 9 is 3/4

81/16 divided by 27/4 = 3/4

243/64 divided by 81/16 = 3/4

So r = 3/4

This is a geometric series with |r| < 1
so it converges, and we can evaluate what
it converges to.  Use the formula:

       a1
S = ———————
     1 - r 

a1 = 1st term = 9
r = 3/4

        9
S = —————————
     1 - 3/4 
 
        9
S = —————————
       1/4

S = 36

So the series converges to 36, and this is called its
evaluation.

Edwin
AnlytcPhil@aol.com
Answer by astromathman(21)   (Show Source): You can put this solution on YOUR website!
I am (initially) assuming the question is supposed to represent an infinite series:
9 + 27/4 + 81/16 + 243/64 + ...
Note that this is a geometric series. The common ratio is 3/4. (You can see this because the numerators are successive multiples of 3 and the denominators are successive multiples of 4.) Since the absolute value of the common ratio is less than 1, the series converges.
The sum of an infinite geometric series is a/(1-r), where a is the first term and r is the common ratio, so the sum is 9/(1-(3/4)) = 9/(1/4) = 9*4 = 36.
If the problem REALLY IS the finite sum of these four terms, the formula is a(1-r^n)/(1-r), where n is the term-number of the last term, 4 in this case.
Interestingly, the formula for the infinite sum is easier to remember than the formula for the finite sum! To be able to do finite sums knowing only the infinite sum formula, find the infinite sum starting with the first term (9), then find the infinite sum starting with the first "missing" term (i.e. the last stated term times the common ratio: (243/64)*(3/4)), then subtract. The answer (rounded to the third decimal place) is 24.609.
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