What is value the of `5/(sqrt2+1)+5/(sqrt3+sqrt2)+5/(sqrt4+sqrt3)+….. 5/(sqrt121+sqrt120)?`
(a)25
(b)50
(c)100
(d)75

Options

A. 25
B. 50
C. 100
D. 75

Answer

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

\[ S = \frac{5}{\sqrt{2}+1} + \frac{5}{\sqrt{3}+\sqrt{2}} + \frac{5}{\sqrt{4}+\sqrt{3}} + \ldots + \frac{5}{\sqrt{121}+\sqrt{120}} \]

### Step 1: Factor out the common term
Since each term in the sum has a common factor of 5 in the numerator, we can factor it out:

\[ S = 5 \left( \frac{1}{\sqrt{2}+1} + \frac{1}{\sqrt{3}+\sqrt{2}} + \frac{1}{\sqrt{4}+\sqrt{3}} + \ldots + \frac{1}{\sqrt{121}+\sqrt{120}} \right) \]

### Step 2: Simplify each term using conjugates
For each term of the form \(\frac{1}{\sqrt{n} + \sqrt{n-1}}\), we can multiply the numerator and the denominator by the conjugate \(\sqrt{n} - \sqrt{n-1}\):

\[ \frac{1}{\sqrt{n} + \sqrt{n-1}} \cdot \frac{\sqrt{n} - \sqrt{n-1}}{\sqrt{n} - \sqrt{n-1}} = \frac{\sqrt{n} - \sqrt{n-1}}{(\sqrt{n})^2 - (\sqrt{n-1})^2} = \frac{\sqrt{n} - \sqrt{n-1}}{n - (n-1)} = \sqrt{n} - \sqrt{n-1} \]

### Step 3: Rewrite the sum
Now, we can rewrite \(S\):

\[ S = 5 \left( (\sqrt{2} - 1) + (\sqrt{3} - \sqrt{2}) + (\sqrt{4} - \sqrt{3}) + \ldots + (\sqrt{121} - \sqrt{120}) \right) \]

### Step 4: Notice the telescoping nature of the series
The series is telescoping, meaning that most terms will cancel out:

\[ S = 5 \left( \sqrt{121} - 1 \right) \]

### Step 5: Calculate the final value
Since \(\sqrt{121} = 11\):

\[ S = 5 \left( 11 - 1 \right) = 5 \times 10 = 50 \]

### Final Answer
Thus, the value of the expression is:

\[ \boxed{50} \]