The value of `sum_(r=1)^n1/(sqrt(a+r x)+sqrt(a+(r-1)x))` is -

Options

A. `n/(sqrta + sqrt(a+nx))`
B. `(sqrt(a+nx) - sqrta)/n`
C. `(n(sqrt(a+nx)-a))/x`
D. `(xsqrt(a+nx)-sqrta)/x`

Answer

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

\[ S = \sum_{r=1}^{n} \frac{1}{\sqrt{a + rx} + \sqrt{a + (r-1)x}} \]

### Step 1: Rationalizing the Denominator
We start by rationalizing the denominator. We can multiply the numerator and the denominator by \(\sqrt{a + rx} - \sqrt{a + (r-1)x}\):

\[ S = \sum_{r=1}^{n} \frac{\sqrt{a + rx} - \sqrt{a + (r-1)x}}{(\sqrt{a + rx} + \sqrt{a + (r-1)x})(\sqrt{a + rx} - \sqrt{a + (r-1)x})} \]

### Step 2: Simplifying the Denominator
The denominator simplifies as follows:

\[ (\sqrt{a + rx})^2 - (\sqrt{a + (r-1)x})^2 = (a + rx) - (a + (r-1)x) = x \]

Thus, we can rewrite \(S\):

\[ S = \sum_{r=1}^{n} \frac{\sqrt{a + rx} - \sqrt{a + (r-1)x}}{x} \]

### Step 3: Factoring Out \(\frac{1}{x}\)
Now, we can factor out \(\frac{1}{x}\) from the summation:

\[ S = \frac{1}{x} \sum_{r=1}^{n} \left( \sqrt{a + rx} - \sqrt{a + (r-1)x} \right) \]

### Step 4: Recognizing the Telescoping Series
Notice that this is a telescoping series. Most terms will cancel out:

\[ S = \frac{1}{x} \left( \sqrt{a + nx} - \sqrt{a} \right) \]

### Final Result
Thus, the value of the summation \(S\) is:

\[ S = \frac{\sqrt{a + nx} - \sqrt{a}}{x} \]

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