Okay, so I need to find the 2003rd term in the sequence of odd numbers. Let me think. The sequence given is 1, 3, 5, 7, and so on. Hmm, that's the sequence of odd numbers starting from 1. I remember that the nth term of such a sequence can be found using a formula. Let me recall. 

First, let's write down the first few terms and see if there's a pattern. The first term is 1, which is 2*1 - 1. The second term is 3, which is 2*2 - 1. The third term is 5, which is 2*3 - 1. Wait, so the nth term seems to be 2n - 1. Let me check that. For n=1, 2*1 -1 =1. Correct. For n=2, 2*2 -1=3. Correct. For n=3, 2*3 -1=5. Correct. So the formula for the nth term is 2n -1. 

Therefore, to find the 2003rd term, I just need to plug in n=2003 into the formula. Let me calculate that. 2*2003 = 4006. Then subtract 1: 4006 -1 =4005. So the 2003rd term should be 4005. 

Wait, let me double-check. If I take n=1, 2*1 -1=1. Correct. n=2, 3. Correct. So the formula works. Therefore, for n=2003, it's 2*2003 -1. Let me compute that again. 2003 multiplied by 2 is 4006, subtract 1 gives 4005. That seems right. 

Alternatively, maybe I can think of it as the sequence starts at 1 and each term increases by 2. So the first term is 1, the second term is 1 + 2*1, the third term is 1 + 2*2, and so on. So the nth term is 1 + 2*(n-1). Let's check that. For n=1, 1 + 2*0=1. Correct. For n=2, 1 +2*1=3. Correct. For n=3, 1 +2*2=5. Correct. So that's another way to express the same formula. Let's see if these two formulas are equivalent. 

1 + 2*(n-1) = 1 + 2n -2 = 2n -1. Yes, same as before. So both formulas are consistent. Therefore, using either formula, plugging in n=2003 will give 2*2003 -1 =4005. 

So I think that's the answer. But let me just make sure I didn't make a calculation error. Let's compute 2003*2. 2000*2=4000, and 3*2=6, so 4000+6=4006. Then subtract 1 gives 4005. Correct. 

Alternatively, maybe I can use another approach. The number of terms in the sequence up to term k is k. So the first term is 1, second 3, etc. The difference between consecutive terms is 2. So it's an arithmetic sequence with first term a1=1 and common difference d=2. The nth term of an arithmetic sequence is a1 + (n-1)*d. So plugging in, 1 + (n-1)*2 =2n -2 +1=2n -1. Same formula again. So that's consistent. 

Therefore, regardless of the method, the formula for the nth term is 2n -1. Therefore, the 2003rd term is 2*2003 -1=4005. 

I think that's solid. Let me check with a smaller n. For example, n=4. The 4th term should be 7. Plugging into the formula: 2*4 -1=8-1=7. Correct. n=5: 2*5 -1=10-1=9. Correct. So formula works. 

Alternatively, if I consider that the sequence of odd numbers is all numbers of the form 2k -1 where k is a positive integer. So k=1 gives 1, k=2 gives 3, etc. Therefore, the 2003rd term would be 2*2003 -1=4005. 

Yes, that's the same as before. So all approaches lead to the same result. Therefore, I can be confident that the answer is 4005. 

I think that's thorough. I considered multiple ways to derive the formula, checked with smaller numbers, and confirmed that the calculation is correct. No steps seem to be missing or errors in reasoning. So the final answer should be 4005.
Therefore, the 2003rd term in the sequence of odd numbers is $\boxed{4005}$.
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