SURE
Okay, so I need to figure out how to solve this problem where I have to find the sum of the first 100 terms of the sequence defined by a_n = 2n - 1. Let me start by understanding what the sequence looks like.
First, let me write out the first few terms to see the pattern. The nth term is given by a_n = 2n - 1. So for n=1, a_1 = 21 - 1 = 1. For n=2, a_2 = 22 - 1 = 3. Then a_3 = 5, a_4 = 7, and so on. So this is the sequence of odd numbers: 1, 3, 5, 7, 9, etc. That makes sense because each term increases by 2, which is the definition of an arithmetic sequence with common difference 2.
Now, the problem is to find the sum of the first 100 terms of this sequence. Since it's an arithmetic sequence, I remember there's a formula for the sum of the first n terms. Let me recall that formula. I think it's S_n = n/2 * (a_1 + a_n), where S_n is the sum, n is the number of terms, a_1 is the first term, and a_n is the nth term. Alternatively, sometimes it's written as S_n = n/2 * , where d is the common difference. Either formula should work, but I need to make sure I use the right one based on the information I have.
In this case, I know a_1 is 1, and the common difference d is 2. But I also need to find a_100, the 100th term, to use the first formula. Let me compute a_100 first. Using the formula for the nth term of an arithmetic sequence: a_n = a_1 + (n - 1)d. Plugging in the values: a_100 = 1 + (100 - 1)2 = 1 + 992 = 1 + 198 = 199. So the 100th term is 199.
https://www.18122.com/url/usdt/usdt4933.jpg
Now, using the first sum formula: S_100 = 100/2 * (1 + 199) = 50 * 200 = 10,000. Wait, that seems straightforward. Let me check with the other formula to verify. The second formula is S_n = n/2 * . Plugging in the values: S_100 = 100/2 * = 50 * = 50 * 200 = 10,000. Same result. That's a good sign.
But let me think again to make sure I didn't make a mistake. The sequence is 1, 3, 5, ..., 199. The sum of an arithmetic sequence can also be calculated by averaging the first and last term and multiplying by the number of terms. So (1 + 199)/2 = 200/2 = 100, and then 100 terms times 100 gives 10,000. Yep, that checks out.
Alternatively, I remember that the sum of the first n odd numbers is n². Wait, is that true? Let me test with small n. For n=1: 1 = 1². For n=2: 1 + 3 = 4 = 2². For n=3: 1 + 3 + 5 = 9 = 3². Oh, that's a neat formula! So if the sum of the first n odd numbers is n², then for n=100, the sum should be 100² = 10,000. That's the same answer as before. So three different methods all give 10,000. That gives me confidence that the answer is correct.
But just to be thorough, let me go through the steps again. First term is 1, common difference 2. Number of terms is 100. Using the formula S_n = n/2 . Plugging in: 100/2 = 50 = 50200 = 10,000. Yep. Alternatively, using the formula with a_n: a_100 = 1 + 992 = 199. Then S_100 = 100*(1 + 199)/2 = 100200/2 = 100100 = 10,000. And the sum of first n odd numbers is n², so 100² = 10,000. All methods converge to the same result.
I think that's solid. But let me just consider if there's any chance I messed up the value of a_100. Let's compute a_100 again step by step. The nth term is 2n - 1. So for n=100: 2*100 = 200, minus 1 is 199. Correct. So a_100 is definitely 199. Then adding 1 and 199 gives 200, divided by 2 is 100, multiplied by 100 terms gives 10,000. Perfect.
Another way to think about it: the sequence of odd numbers. The sum of the first n odd numbers is a well-known formula that equals n squared. Since 100 is a perfect square, 100 squared is 10,000. This is a quicker way if you remember that property. But even if I didn't remember that, the arithmetic series formulas work perfectly here.
I guess the key steps here are recognizing that the sequence is arithmetic with a common difference of 2, then applying the appropriate formula for the sum. Alternatively, recognizing the pattern of odd numbers and using the sum formula for that specific case. Either way, the answer is 10,000. I don't see any mistakes in the reasoning, and multiple methods confirm the same result. So I'm pretty confident that the sum of the first 100 terms is 10,000.
The sequence ( a_n = 2n - 1 ) generates the first 100 odd numbers: 1, 3, 5, ..., 199. To find the sum of the first 100 terms:
Identify the sequence type: This is an arithmetic sequence with first term ( a_1 = 1 ) and common difference ( d = 2 ).
Find the 100th term:
[
a_{100} = 2(100) - 1 = 199
]
Use the sum formula for an arithmetic series:
[
S_n = \frac{n}{2} \times (a_1 + a_n)
]
Substituting ( n = 100 ), ( a_1 = 1 ), and ( a_{100} = 199 ):
[
S_{100} = \frac{100}{2} \times (1 + 199) = 50 \times 200 = 10,!000
]
Alternative verification using the sum of first ( n ) odd numbers:
The sum of the first ( n ) odd numbers is ( n^2 ). For ( n = 100 ):
[
S_{100} = 100^2 = 10,!000
]
Final Answer: (\boxed{10000})
页:
[1]