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\(\sum_{n=3}^{7} 2n+3\)

Guest Aug 4, 2017
 #1
avatar
0

Is it "Geometric Series" ??

9, 11, 13, 15, 17 =65 ??!!.

Guest Aug 4, 2017
 #2
avatar+2248 
+1

To do this problem, evaluate n at the values from the positive integers n=3 to n=7

 

When n=3:

\(2n+3\) Plug in the appropriate value for n, which is 3, in this case.
\(2*3+3\)  
\(6+3\)  
\(9\)  
   

 

Now, let's evaluate it for n=4:

 

\(2n+3\) Plug in n=4
\(2*4+3\)  
\(8+3\)  
\(11\)  
   

 

n=5:

\(2n+3\) Replace all instances of n with 5
\(2*5+3\)  
\(10+3\)  
\(13\)  
   

 

n=6:

\(2n+3\) Plug in 6 for n
\(2*6+3\)  
\(12+3\)  
\(15\)  
   

 

Now, I do not have to evaluate 7 as I can infer, based on the information above, that the next value in the series will be the sum of the last number in the series and 2. Therefore, 2n+3 when n=7, the expression evaluates to 17.

 

Now, our last step is too add the series together:

 

\(9+11+13+15+17\) Figure out what this evaluates to. Of course, with addition you can add in whatever order you'd like. I'll use this property to my advantage to ease calculation.
\(20+13+15+17\)  
\(20+30+15\)  
\(50+15\)  
\(65\)  
   
TheXSquaredFactor  Aug 4, 2017
 #3
avatar+90088 
+1

 

Here's another way :

 

The first and last terms sum to 26....so....the sum is given by

 

[ Sum of first and last term] [ Indices of last term  - indices of first term + 1 ]  / 2  =

 

[26 ] [7 - 3 + 1 ] / 2  =

 

[26] [ 5] / 2  =

 

[26 / 2 ] * 5   =

 

13 * 5  =

 

65

 

 

cool cool cool

CPhill  Aug 4, 2017

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