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What is ∑n=15[5(−2)^n−1] equal to?

Guest May 17, 2018
edited by Guest  May 17, 2018
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Although the notation is sloppy, I will assume based on former questions of the same that "∑n=15[5(−2)^n−1]" translates mathematically to \(\sum_{n=1}^5[5(-2)^{n-1}]\). Please correct me, though, if this in incorrect. 

 

Anyway, there is neat formula that will deal with question very quickly. It is the following:

 

\(S_n=a_1\left(\frac{1-r^n}{1-r}\right),r\neq1\)

 

Sis the sum of the geometric series

ais the first term in the series

r is the common ratio

n is the number of terms in the finite geometric series

 

Let's use this formula to our advantage! However, there are two variables that we do not know: aand  n. Let's start with a1.The first term in the series is when n=1, so let's determine what that is. 
 

\(a_1=5(-2)^{1-1}\) A lot of simplification will occur here--especially since the exponent simplifies to zero. 
\(a_1=5\)  
   

 

The number of terms also must be calculated:

 

\(n=5-1+1\) To find the number of terms, subtract the maximum n, 5, from the minimum, one. Add one to make up for the missing one.
\(n=5\) We can do the formula!
   

 

\(a_1=5,n=5,r=-2;\\ S_n=a_1\left(\frac{1-r^n}{1-r}\right)\) Do the substitution!
\(S_n=5\left(\frac{1-(-2)^5}{1-(-2)}\right)\) Let's simplify the numerator and denominator.
\(S_n=5\left(\frac{1-(-32)}{1+2}\right)\)  
\(S_n=5\left(\frac{33}{3}\right)\)  
\(S_n=5*11=55\) Just like that, we are done!
   
 
TheXSquaredFactor  May 17, 2018

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