If `S_(lambda) = sum_(r = 0)^(oo) (1)/(lambda^(r)),"then" sum_(lambda = 1)^(n) (lambda - 1)S_(lambda) =`

Options

A. `(n(n-1))/(2)`
B. `(n(n + 1))/(2)`
C. `(n (n + 2))/(2)`
D. none

Answer

To solve the problem, we need to evaluate the expression:

\[ \sum_{\lambda = 1}^{n} (\lambda - 1) S_{\lambda} \]

where

\[ S_{\lambda} = \sum_{r = 0}^{\infty} \frac{1}{\lambda^r} \]

### Step 1: Evaluate \( S_{\lambda} \)

The series \( S_{\lambda} \) is a geometric series. The first term \( a \) is 1 (when \( r = 0 \)) and the common ratio \( r \) is \( \frac{1}{\lambda} \).

The formula for the sum of an infinite geometric series is:

\[ S = \frac{a}{1 - r} \]

Substituting the values:

\[ S_{\lambda} = \frac{1}{1 - \frac{1}{\lambda}} = \frac{1}{\frac{\lambda - 1}{\lambda}} = \frac{\lambda}{\lambda - 1} \]

### Step 2: Substitute \( S_{\lambda} \) into the main expression

Now, substituting \( S_{\lambda} \) back into the summation:

\[ \sum_{\lambda = 1}^{n} (\lambda - 1) S_{\lambda} = \sum_{\lambda = 1}^{n} (\lambda - 1) \left( \frac{\lambda}{\lambda - 1} \right) \]

This simplifies to:

\[ \sum_{\lambda = 1}^{n} \lambda \]

### Step 3: Evaluate the summation \( \sum_{\lambda = 1}^{n} \lambda \)

The sum of the first \( n \) natural numbers is given by the formula:

\[ \sum_{\lambda = 1}^{n} \lambda = \frac{n(n + 1)}{2} \]

### Final Result

Thus, the final result for the expression is:

\[ \sum_{\lambda = 1}^{n} (\lambda - 1) S_{\lambda} = \frac{n(n + 1)}{2} \]