Let `S_(k)=underset(ntooo)limunderset(i=0)overset(n)sum(1)/((k+1)^(i))." Then "underset(k=1)overset(n)sumkS_(k)` equals

Options

A. `(n (n +1))/(2)`
B. `(n (n -1))/(2)`
C. `(n (n +2))/(2)`
D. `(n (n +3))/(2)`

Answer

To solve the problem step by step, we need to evaluate the expression \( \sum_{k=1}^{n} k S_k \), where \( S_k = \lim_{n \to \infty} \sum_{i=0}^{n} \frac{1}{(k+1)^i} \).

### Step 1: Evaluate \( S_k \)

We start by evaluating \( S_k \):
\[ S_k = \lim_{n \to \infty} \sum_{i=0}^{n} \frac{1}{(k+1)^i} \]
This is a geometric series where the first term \( a = 1 \) and the common ratio \( r = \frac{1}{k+1} \).

The sum of an infinite geometric series is given by:
\[ \text{Sum} = \frac{a}{1 - r} \]
Here, \( a = 1 \) and \( r = \frac{1}{k+1} \). Thus,
\[ S_k = \frac{1}{1 - \frac{1}{k+1}} = \frac{1}{\frac{k+1-1}{k+1}} = \frac{k+1}{k} \]

### Step 2: Substitute \( S_k \) into the summation

Now, we substitute \( S_k \) into the expression \( \sum_{k=1}^{n} k S_k \):
\[ \sum_{k=1}^{n} k S_k = \sum_{k=1}^{n} k \cdot \frac{k+1}{k} = \sum_{k=1}^{n} (k + 1) \]

### Step 3: Simplify the summation

We can separate the summation:
\[ \sum_{k=1}^{n} (k + 1) = \sum_{k=1}^{n} k + \sum_{k=1}^{n} 1 \]
The first part, \( \sum_{k=1}^{n} k \), is the sum of the first \( n \) natural numbers, which is given by:
\[ \sum_{k=1}^{n} k = \frac{n(n+1)}{2} \]
The second part, \( \sum_{k=1}^{n} 1 \), simply counts the number of terms, which is \( n \).

### Step 4: Combine the results

Now we combine the results:
\[ \sum_{k=1}^{n} (k + 1) = \frac{n(n+1)}{2} + n \]
To combine these, we can express \( n \) as \( \frac{2n}{2} \):
\[ \sum_{k=1}^{n} (k + 1) = \frac{n(n+1)}{2} + \frac{2n}{2} = \frac{n(n+1) + 2n}{2} = \frac{n^2 + n + 2n}{2} = \frac{n^2 + 3n}{2} \]

### Step 5: Final expression

Thus, we have:
\[ \sum_{k=1}^{n} k S_k = \frac{n(n + 3)}{2} \]

### Final Answer
The final answer is:
\[ \frac{n(n + 3)}{2} \]