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 solve the problem, we need to find the sum of the series:

\[ S = \sum_{n=1}^{24} \frac{1}{n \sqrt{n} + (n-1) \sqrt{n+1}} \]

### Step 1: Rewrite the General Term
The general term of the series can be expressed as:

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

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

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

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

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

### Step 4: Simplify the Denominator
Calculating the denominator:

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

Expanding both terms:

\[ = n^3 + n^2 - (n^2 - 2n + 1)n = n^3 + n^2 - n^3 + 2n^2 - n = 3n^2 - n \]

### Step 5: Substitute Back into the General Term
Thus, we have:

\[ T_n = \frac{n \sqrt{n+1} - (n-1) \sqrt{n}}{3n^2 - n} \]

### Step 6: Rewrite the Series
Now we can rewrite the series \(S\):

\[ S = \sum_{n=1}^{24} \left( \frac{n \sqrt{n+1} - (n-1) \sqrt{n}}{3n^2 - n} \right) \]

### Step 7: Simplify Further
Notice that the series has a telescoping nature. We can express the series as:

\[ S = \sum_{n=1}^{24} \left( \frac{1}{\sqrt{n}} - \frac{1}{\sqrt{n+1}} \right) \]

### Step 8: Evaluate the Series
This is a telescoping series, where most terms cancel out:

\[ S = \left( 1 - \frac{1}{\sqrt{25}} \right) = 1 - \frac{1}{5} = \frac{4}{5} \]

### Final Answer
Thus, the sum of the series is:

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