The sum of ` (1)/( 2sqrt1+1 sqrt2 ) + (1)/( 3 sqrt2 + 2 sqrt3 ) + (1)/( 4 sqrt3 + 3 sqrt4 ) +...+(1)/( 25 sqrt 24 + 24 sqrt25)` is

Answer

To find the sum of the series

\[ S = \frac{1}{2\sqrt{1} + 1\sqrt{2}} + \frac{1}{3\sqrt{2} + 2\sqrt{3}} + \frac{1}{4\sqrt{3} + 3\sqrt{4}} + \ldots + \frac{1}{25\sqrt{24} + 24\sqrt{25}}, \]

we will first identify the general term of the series and then simplify it.

### Step 1: Identify the General Term
The general term \( T_n \) can be expressed as:

\[ T_n = \frac{1}{n\sqrt{n-1} + (n-1)\sqrt{n}}. \]

### Step 2: Rationalize the Denominator
To simplify \( T_n \), we will rationalize the denominator. We multiply the numerator and the denominator by the conjugate of the denominator:

\[ T_n = \frac{1}{n\sqrt{n-1} + (n-1)\sqrt{n}} \cdot \frac{n\sqrt{n-1} - (n-1)\sqrt{n}}{n\sqrt{n-1} - (n-1)\sqrt{n}}. \]

### Step 3: Apply the Difference of Squares
Using the difference of squares formula \( a^2 - b^2 \), we can simplify the denominator:

\[ (n\sqrt{n-1})^2 - ((n-1)\sqrt{n})^2 = n^2(n-1) - (n-1)^2n = n(n-1)(n - (n-1)) = n(n-1). \]

### Step 4: Simplify the Numerator
The numerator becomes:

\[ n\sqrt{n-1} - (n-1)\sqrt{n}. \]

So we have:

\[ T_n = \frac{n\sqrt{n-1} - (n-1)\sqrt{n}}{n(n-1)}. \]

### Step 5: Separate the Terms
Now we can separate the terms in the numerator:

\[ T_n = \frac{1}{n\sqrt{n}} - \frac{1}{(n-1)\sqrt{n-1}}. \]

### Step 6: Sum the Series
Now, we can sum the series from \( n = 2 \) to \( n = 25 \):

\[ S = \sum_{n=2}^{25} \left( \frac{1}{n\sqrt{n}} - \frac{1}{(n-1)\sqrt{n-1}} \right). \]

### Step 7: Telescoping Series
This is a telescoping series. Most terms will cancel out:

\[ S = \left( \frac{1}{2\sqrt{2}} - \frac{1}{1\sqrt{1}} \right) + \left( \frac{1}{3\sqrt{3}} - \frac{1}{2\sqrt{2}} \right) + \ldots + \left( \frac{1}{25\sqrt{25}} - \frac{1}{24\sqrt{24}} \right). \]

After cancellation, we are left with:

\[ S = \frac{1}{25} - 1. \]

### Step 8: Final Calculation
Calculating the final result gives:

\[ S = 1 - \frac{1}{5} = \frac{4}{5}. \]

Thus, the sum of the series is:

\[ \boxed{\frac{4}{5}}. \]