QUESTION 9 he sum of $$ n $$ terms is aiven by $$ S_{n}=\frac{n}{( }(1+n) $$, find $$ T_{ ext {. }} $$

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QUESTION 9 he sum of $$ n $$ terms is aiven by $$ S_{n}=\frac{n}{( }(1+n) $$, find $$ T_{	ext {. }} $$

UpStudy AI · Accepted Answer

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:**

$$
T_n = \frac{1}{n(n + 1)}
$$
The $$ n $$-th term $$ T_n $$ is $$ \frac{1}{n(n + 1)} $$.

QUESTION 9 he sum of \( n \) terms is aiven by \( S_{n}=\frac{n}{( }(1+n) \), find \( T_{\text {. }} \)

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