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

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, \[ T_n = \frac{1}{n(1+n)} \]