SC
Okay, the user just sent "SC" as a query. That's pretty minimal. Hmm, I wonder what they're really looking for. Could be an abbreviation for many things - maybe they're in a hurry or typing from a mobile device.
First thought: Are they referring to South Carolina? That's the most common meaning of "SC" in general contexts. But could also be something else entirely - like a school subject, a company, or even a typo.
I should cover the most probable meanings first. South Carolina makes sense as a starting point since it's widely recognized. But I'll also include other common interpretations like "Senior Consultant" (business context) and "Santa Claus" (holiday season context).
The user didn't provide any additional clues, so I'll list possibilities with brief explanations. If they need more details, they can clarify. Better to be thorough than assume.
...Wait, is this for a school assignment? The all-caps makes me think it could be an abbreviation test. But no way to know for sure. I'll keep explanations simple and factual.
Noticing they used English, so I'll respond in English. No need to overcomplicate - just present options neutrally. If they wanted something specific, they'll probably follow up.
Final decision: List 4-5 most likely meanings with one-sentence explanations each. Keep it scannable. If they need depth, they'll ask.
The abbreviation "SC" has several common meanings depending on the context. Here are the most frequent interpretations:
South Carolina: A state in the southeastern United States (US postal abbreviation).
Senior Consultant: A common job title in professional services firms (like consulting, accounting, law).
Santa Claus: Often used informally, especially around Christmas time.
Service Center: A location providing specific services (e.g., call center, support center, medical service center).
Standard Chartered: A major international banking group (ticker symbol on stock exchanges).
Sickle Cell: Referring to sickle cell disease or sickle cell trait.
Subcutaneous: A medical term meaning "under the skin" (e.g., SC injection).
Supreme Court: Referring to the highest court in a country or state (e.g., US Supreme Court, South Carolina Supreme Court).
StarCraft: A popular real-time strategy video game.
Scientific Committee: A group established to provide scientific advice on a specific topic.
To get the most accurate answer, please provide more context! For example:
Are you filling out a form? (Location? Job title?)
Reading a document? (Financial report? Medical chart? News article?)
Playing a game? (Video game?)
Talking about Christmas? (Holiday planning?)
Without context, "South Carolina" is often the most likely meaning.
Okay, so I need to figure out how to solve this problem where I have to find the value of S, which is the sum from n=1 to infinity of 1/(n(n+1)(n+2)). Hmm, let's see. I remember that sums involving reciprocals of polynomial terms can sometimes be tackled using partial fractions. Maybe I should start by breaking down the general term into simpler fractions.
The term is 1/(n(n+1)(n+2)). Let me try to express this as A/n + B/(n+1) + C/(n+2). To find A, B, and C, I need to combine these fractions and equate the numerators. Let's do that step by step.
First, write the equation: 1 = A(n+1)(n+2) + Bn(n+2) + Cn(n+1). Now, I need to find constants A, B, and C that satisfy this equation for all n. To find these constants, I can substitute specific values of n that simplify the equation.
Let's start with n = 0. Plugging in n=0, we get 1 = A(1)(2) + B(0) + C(0), so 1 = 2A, which means A = 1/2.
Next, let's substitute n = -1. Then, the equation becomes 1 = A(0)(1) + B(-1)(1) + C(-1)(0). Simplifying, 1 = -B, so B = -1.
Then, substitute n = -2. Plugging in n = -2, we get 1 = A(-1)(0) + B(-2)(0) + C(-2)(-1). That simplifies to 1 = 2C, so C = 1/2.
So, the partial fractions decomposition is 1/(2n) - 1/(n+1) + 1/(2(n+2)). Let me check if that works. Let's combine them again:
1/(2n) - 1/(n+1) + 1/(2(n+2)) = [ (n+1)(n+2) - 2n(n+2) + n(n+1) ] / .
Calculating the numerator:
First term: (n+1)(n+2) = n² + 3n + 2
Second term: -2n(n+2) = -2n² -4n
Third term: n(n+1) = n² + n
Adding them up: (n² +3n +2) + (-2n² -4n) + (n² +n) = (n² -2n² +n²) + (3n -4n +n) + 2 = 0 + 0 + 2 = 2
So the numerator is 2, and the denominator is 2n(n+1)(n+2), so overall it's 2/ = 1/, which matches the original term. Great, so the partial fractions are correct.
Now, the sum S becomes the sum from n=1 to infinity of . Let's split this into three separate sums:
https://www.18122.com/url/usdt/usdt0239.JPEG
S = (1/2)Σ_{n=1}^∞ 1/n - Σ_{n=1}^∞ 1/(n+1) + (1/2)Σ_{n=1}^∞ 1/(n+2)
Hmm, but these sums are telescoping, right? Let's adjust the indices to see if terms cancel out.
First sum: Σ_{n=1}^∞ 1/n is the harmonic series, which diverges, but when combined with the other sums, maybe they cancel out.
Second sum: Σ_{n=1}^∞ 1/(n+1) = Σ_{k=2}^∞ 1/k where k = n+1.
Third sum: Σ_{n=1}^∞ 1/(n+2) = Σ_{m=3}^∞ 1/m where m = n+2.
So rewriting S:
S = (1/2)(Σ_{n=1}^∞ 1/n) - (Σ_{k=2}^∞ 1/k) + (1/2)(Σ_{m=3}^∞ 1/m)
Now, let's express all sums starting from the same index. Let's write each sum in terms of Σ_{n=1}^∞ 1/n.
First term is already (1/2)Σ_{n=1}^∞ 1/n.
Second term is Σ_{k=2}^∞ 1/k = Σ_{n=1}^∞ 1/n - 1/1.
Third term is Σ_{m=3}^∞ 1/m = Σ_{n=1}^∞ 1/n - 1/1 - 1/2.
So substituting back into S:
S = (1/2)Σ_{n=1}^∞ 1/n - [Σ_{n=1}^∞ 1/n - 1] + (1/2)[Σ_{n=1}^∞ 1/n - 1 - 1/2]
Let me expand this:
S = (1/2)Σ1/n - Σ1/n + 1 + (1/2)Σ1/n - (1/2)(1) - (1/2)(1/2)
Combine like terms:
The Σ1/n terms:
(1/2)Σ1/n - Σ1/n + (1/2)Σ1/n = (1/2 -1 +1/2)Σ1/n = 0Σ1/n. So the divergent harmonic series terms cancel out!
Then the constants:
1 - (1/2) - (1/4) = 1 - 3/4 = 1/4.
So S = 1/4.
Wait, that seems too straightforward. Let me verify by computing partial sums.
Let's compute the partial sum up to N terms and see if it approaches 1/4.
The partial sum S_N = Σ_{n=1}^N
Let's split this into three sums:
S_N = (1/2)Σ_{n=1}^N 1/n - Σ_{n=1}^N 1/(n+1) + (1/2)Σ_{n=1}^N 1/(n+2)
Adjusting indices:
First sum: (1/2)(1 + 1/2 + 1/3 + ... + 1/N)
Second sum: (1/2 + 1/3 + ... + 1/(N+1))
Third sum: (1/2)(1/3 + 1/4 + ... + 1/(N+2))
So combining these:
S_N = (1/2)(1 + 1/2 + 1/3 + ... + 1/N) - (1/2 + 1/3 + ... + 1/(N+1)) + (1/2)(1/3 + ... + 1/(N+2))
Let's write all terms:
First part: 1/2 + 1/4 + 1/6 + ... + 1/(2N)
Second part: -1/2 -1/3 - ... -1/(N+1)
Third part: 1/6 + 1/8 + ... + 1/(2(N+2))
Wait, maybe another approach. Let's write S_N as:
S_N = (1/2)(H_N) - (H_{N+1} -1) + (1/2)(H_{N+2} -1 -1/2)
Where H_N is the Nth harmonic number.
Expanding this:
S_N = (1/2)H_N - H_{N+1} +1 + (1/2)H_{N+2} - (1/2)(3/2)
Simplify:
= (1/2)H_N - (H_N + 1/(N+1)) +1 + (1/2)(H_N + 1/(N+1) + 1/(N+2)) - 3/4
Combine terms:
= (1/2)H_N - H_N -1/(N+1) +1 + (1/2)H_N + (1/(2(N+1)) + 1/(2(N+2))) - 3/4
Simplify:
The H_N terms: (1/2 -1 +1/2)H_N = 0
The 1/(N+1) terms: -1/(N+1) +1/(2(N+1)) = -1/(2(N+1))
The 1/(N+2) term: +1/(2(N+2))
The constants: 1 - 3/4 = 1/4
So S_N = 1/4 -1/(2(N+1)) +1/(2(N+2))
Combine the last two terms:
=1/4 + [ -1/(2(N+1)) +1/(2(N+2)) ] =1/4 + [ (-(N+2) + (N+1)) / (2(N+1)(N+2)) ]
Simplify numerator:
-(N+2) + (N+1) = -N -2 +N +1 = -1
So S_N =1/4 -1/(2(N+1)(N+2))
Therefore, as N approaches infinity, the term -1/(2(N+1)(N+2)) approaches 0, so the limit is 1/4.
Yes, that matches the previous result. Therefore, the sum S converges to 1/4.
To find the value of ( S = \sum_{n=1}^{\infty} \frac{1}{n(n+1)(n+2)} ), we use partial fraction decomposition:
[
\frac{1}{n(n+1)(n+2)} = \frac{1}{2n} - \frac{1}{n+1} + \frac{1}{2(n+2)}
]
This decomposition is verified by combining the fractions and confirming the numerator equals 1. Substituting this into the series:
[
S = \sum_{n=1}^{\infty} \left( \frac{1}{2n} - \frac{1}{n+1} + \frac{1}{2(n+2)} \right)
]
Breaking into three separate sums and adjusting indices:
[
S = \frac{1}{2} \sum_{n=1}^\infty \frac{1}{n} - \sum_{n=2}^\infty \frac{1}{n} + \frac{1}{2} \sum_{n=3}^\infty \frac{1}{n}
]
Expressing each sum in terms of the harmonic series ( H_N ):
[
S = \frac{1}{2} H_\infty - (H_\infty - 1) + \frac{1}{2} (H_\infty - 1 - \frac{1}{2})
]
Simplifying, the divergent harmonic series terms cancel, leaving:
[
S = 1 - \frac{3}{4} = \frac{1}{4}
]
Verification via partial sums:
For partial sum ( S_N ):
[
S_N = \frac{1}{4} - \frac{1}{2(N+1)(N+2)}
]
As ( N \to \infty ), the correction term vanishes, confirming ( S = \frac{1}{4} ).
Answer:
\boxed{\dfrac{1}{4}}
页:
[1]