Finding the first 40 positive odd integers

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SUMMARY

The sum of the first 40 positive odd integers is calculated as 1 + 3 + 5 + ... + 77 + 79. The 40th positive odd integer is determined using the formula for the N-th term of an arithmetic sequence, specifically 1 + 2(n - 1). By substituting n with 40, the result is 79, confirming that 79 is indeed the 40th positive odd integer. This method eliminates the need to list all integers sequentially.

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Mr Davis 97
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I have the following problem: What is the sum of the first 40 positive odd integers?

I look at the solution, and it says that "The sum of the first 40 positive odd integers is ##1 + 3 + 5 + \dotsm + 77 + 79##. And then it goes on with the solution.

My question is, how do I find that 79 is the 40th positive odd integer? Do I have to just write them all down until I get to the 40th one?
 
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Mr Davis 97 said:
I have the following problem: What is the sum of the first 40 positive odd integers?

I look at the solution, and it says that "The sum of the first 40 positive odd integers is ##1 + 3 + 5 + \dotsm + 77 + 79##. And then it goes on with the solution.

My question is, how do I find that 79 is the 40th positive odd integer? Do I have to just write them all down until I get to the 40th one?
The first odd positive integer is 1. The second one is 1+2. The third one is 1+2+2. How many times you need to add 2 to get 40 terms?
(What kind of sequence is 1, 3, 5. 7, ... ? What is the N-th term of that sequence?)
 
ehild said:
The first odd positive integer is 1. The second one is 1+2. The third one is 1+2+2. How many times you need to add 2 to get 40 terms?
(What kind of sequence is 1, 3, 5. 7, ... ? What is the N-th term of that sequence?)
Ah, so I can just create an arithmetic sequence ##1 + 2(n - 1)##, where n is the index of the odd number. So when I substitute 40 I just get 1 + 2(39) = 79! Awesome, thanks.
 

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