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# Geometric series

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$$\sum_{n=3}^{7} 2n+3$$

Aug 4, 2017

#1
0

Is it "Geometric Series" ??

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

Aug 4, 2017
#2
+2298
+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. $$20+13+15+17$$ $$20+30+15$$ $$50+15$$ $$65$$
Aug 4, 2017
#3
+94544
+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

Aug 4, 2017