Step-by-step Solution:

the general term of the given series can be written as \( T_n = \frac{1}{(n+1)\sqrt{n} + n\sqrt{n+1}} \).We can factor out \( \sqrt{n(n+1)} \) from the denominator:\( T_n = \frac{1}{\sqrt{n(n+1)}(\sqrt{n+1} + \sqrt{n})} \).Rationalizing the expression by multiplying the numerator and the denominator by \( (\sqrt{n+1} - \sqrt{n}) \), we get:\( T_n = \frac{\sqrt{n+1} - \sqrt{n}}{\sqrt{n(n+1)}((\sqrt{n+1})^2 - (\sqrt{n})^2)} \).\( T_n = \frac{\sqrt{n+1} - \sqrt{n}}{\sqrt{n(n+1)}(n+1 - n)} \).\( T_n = \frac{\sqrt{n+1} - \sqrt{n}}{\sqrt{n(n+1)}} \).Splitting the fraction, we obtain:\( T_n = \frac{1}{\sqrt{n}} - \frac{1}{\sqrt{n+1}} \).The given series is the sum of these terms from \( n = 1 \) to \( n = 24 \).Sum \( = T_1 + T_2 + T_3 + ... + T_{24} \).Sum \( = (\frac{1}{\sqrt{1}} - \frac{1}{\sqrt{2}}) + (\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{3}}) + ... + (\frac{1}{\sqrt{24}} - \frac{1}{\sqrt{25}}) \).This is a telescoping series where all the intermediate terms cancel out.Sum \( = \frac{1}{\sqrt{1}} - \frac{1}{\sqrt{25}} \).Sum \( = 1 - \frac{1}{5} = \frac{4}{5} \).