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how do you calculate \(\displaystyle\sum_{n=1}^{10} (2n + 1)\)??

 Jan 30, 2022
 #1
avatar+675 
+1

\(\displaystyle\sum_{n=1}^{10} \mbox{ } (2n + 1) = (2 \cdot 1 + 1) + (2 \cdot 2 + 1) + \mbox{ }... \mbox{ } + (2 \cdot 10 + 1) = 120\),

 

for example, when n = 2 (the number under the summary sign), then start with 2 till the number over the summary sign. If you don't get it, e.g.

 

\(\displaystyle\sum_{n=2}^{5} \mbox{ } n + 1\) you see n = 2, so (2 + 1), then comes n = 3, so (2 + 1) + (3 + 1) + ... (5 + 1) = 18,

 

so \(\displaystyle\sum_{n=2}^{5} \mbox{ } n + 1 = 18\) .

 

Feel free to ask if you still don't get it.

 Jan 30, 2022
edited by Straight  Jan 30, 2022
 #2
avatar+36417 
+2

a1 =  (2(1))+1 = 3

a10 = 2(10)+ 1 = 21

 

sum =  n/2 ( a1 + a10) = 10/2(3 +21) = 5 * 24 = 120

 Jan 30, 2022
 #3
avatar+675 
0

nice!

Straight  Jan 30, 2022

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