Question
QUESTION 9 he sum of \( n \) terms is aiven by \( S_{n}=\frac{n}{( }(1+n) \), find \( T_{\text {. }} \)
Ask by Hill Dunn. in South Africa
Jan 24,2025
Real Tutor Solution
Tutor-Verified Answer
Answer
The \( n \)-th term \( T_n \) is \( \frac{1}{n(n + 1)} \).
Solution
It appears there might have been a minor typographical error in the expression for the sum of \( n \) terms. Assuming the sum of the first \( n \) terms (\( S_n \)) is given by:
\[
S_n = \frac{n}{n + 1}
\]
We are tasked with finding the \( n \)-th term (\( T_n \)) of the sequence.
### Method to Find \( T_n \):
The \( n \)-th term of a sequence can be found using the difference between the sum of the first \( n \) terms and the sum of the first \( (n-1) \) terms:
\[
T_n = S_n - S_{n-1}
\]
### Calculating \( T_n \):
1. **Expression for \( S_n \):**
\[
S_n = \frac{n}{n + 1}
\]
2. **Expression for \( S_{n-1} \):**
\[
S_{n-1} = \frac{n - 1}{(n - 1) + 1} = \frac{n - 1}{n}
\]
3. **Subtracting to Find \( T_n \):**
\[
T_n = S_n - S_{n-1} = \frac{n}{n + 1} - \frac{n - 1}{n}
\]
4. **Finding a Common Denominator and Simplifying:**
\[
T_n = \frac{n^2 - (n - 1)(n + 1)}{n(n + 1)} = \frac{n^2 - (n^2 - 1)}{n(n + 1)} = \frac{1}{n(n + 1)}
\]
### Final Answer:
\[
T_n = \frac{1}{n(n + 1)}
\]
**Therefore, the \( n \)-th term of the sequence is:**
\[
T_n = \frac{1}{n(n + 1)}
\]
Reviewed and approved by the UpStudy tutoring team
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Simplify this solution Bonus Knowledge
To find \( T_n \), the \( n \)th term of the sequence, we can use the relationship between the sum of the first \( n \) terms \( S_n \) and the \( n \)th term \( T_n \). We know that: \[ T_n = S_n - S_{n-1} \] First, we need to compute \( S_{n-1} \): \[ S_{n-1} = \frac{n-1}{1+(n-1)} = \frac{n-1}{n} \] Now, substitute \( S_n \) and \( S_{n-1} \): \[ S_n = \frac{n}{1+n} \] Then: \[ T_n = S_n - S_{n-1} = \frac{n}{1+n} - \frac{n-1}{n} \] To combine these fractions, find a common denominator: \[ = \frac{n^2 - (1+n)(n-1)}{n(1+n)} = \frac{n^2 - (n^2 - n + n - 1)}{n(1+n)} = \frac{n^2 - (n^2 - 1)}{n(1+n)} = \frac{1}{n(1+n)} \] Thus, \[ T_n = \frac{1}{n(1+n)} \]
