GMAT CLUB OLYMPICS: If 1/5*4^(1/2) +

Official Solution:

If \(\frac{1}{5\sqrt{4} + 4\sqrt{5}} + \frac{1}{6\sqrt{5} + 5\sqrt{6}} +\frac{1}{7\sqrt{6} + 6\sqrt{7}} +...+\frac{1}{n\sqrt{n-1} + (n-1)\sqrt{n}}=\frac{3}{10}\), then what is the value of \(n\)?

A. \(9\)
B. \(10\)
C. \(15\)
D. \(20\)
E. \(25\)


Rationalize by multiplying the denominator and the numerator of each fraction by \(\frac{1}{x\sqrt{x-1} - (x-1)\sqrt{x} }\) (this algebraic manipulation is called rationalization and is performed to eliminate irrational expression in the denominator):

\(\frac{5\sqrt{4} - 4\sqrt{5} }{(5\sqrt{4} + 4\sqrt{5})(5\sqrt{4} - 4\sqrt{5})} + \frac{6\sqrt{5} - 5\sqrt{6} }{(6\sqrt{5} + 5\sqrt{6})(6\sqrt{5} - 5\sqrt{6})} +\frac{7\sqrt{6} - 6\sqrt{7} }{(7\sqrt{6} + 6\sqrt{7})(7\sqrt{6} - 6\sqrt{7})} +...+\frac{n\sqrt{n-1} - (n-1)\sqrt{n} }{(n\sqrt{n-1} + (n-1)\sqrt{n})(n\sqrt{n-1} - (n-1)\sqrt{n})}=\frac{3}{10}\);

\(\frac{5\sqrt{4} - 4\sqrt{5} }{20} + \frac{6\sqrt{5} - 5\sqrt{6} }{30} +\frac{7\sqrt{6} - 6\sqrt{7} }{42} +...+\frac{n\sqrt{n-1} - (n-1)\sqrt{n} }{n(n-1)}=\frac{3}{10}\)

\((\frac{5\sqrt{4} }{20} - \frac{4\sqrt{5} }{20}) + (\frac{6\sqrt{5} }{30} - \frac{5\sqrt{6} }{30}) +(\frac{7\sqrt{6} }{42} - \frac{6\sqrt{7} }{42}) +...+(\frac{n\sqrt{n-1} }{n(n-1)} - \frac{(n-1)\sqrt{n} }{n(n-1)})=\frac{3}{10}\);

\((\frac{\sqrt{4} }{4} - \frac{\sqrt{5} }{5}) + (\frac{\sqrt{5} }{5} - \frac{\sqrt{6} }{6}) +(\frac{\sqrt{6} }{6} - \frac{\sqrt{7} }{7}) +...+(\frac{\sqrt{n-1} }{(n-1)} - \frac{\sqrt{n} }{n})=\frac{3}{10}\);

Notice that everything except the first and the last terms will cancel each other, and we'll be left with \(\frac{\sqrt{4} }{4}- \frac{\sqrt{n} }{n}=\frac{3}{10}\)

\(\frac{1}{2} - \frac{1}{\sqrt{n} } = \frac{3}{10}\);

\(\sqrt{n}=5\);

\(n=25\).


Answer: E­