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If S be the sum of the series, 1/(2√1+1√2)+1/(3√2+2√3)+1/(4√3+3√4)+.....

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S = \(\frac{1}{2\sqrt1+1\sqrt2}\) + \(\frac{1}{3\sqrt2+2\sqrt3}\) + ....................+ \(\frac{1}{36\sqrt35+35\sqrt36}\)

By rationalisation, we get

S = \(\frac{2\sqrt1-1\sqrt2}{4-2}\) + \(\frac{3\sqrt2-2\sqrt3}{18-12}\) + \(\frac{4\sqrt3-3\sqrt4}{48-36}\) + ...............+ \(\frac{36\sqrt{35}-35\sqrt{36}}{45360-44100}\)

\(\frac{2\sqrt1-1\sqrt2}{2}\) + \(\frac{3\sqrt2-2\sqrt3}{6}\) + \(\frac{4\sqrt3-3\sqrt4}{12}\)+............+ \(\frac{36\sqrt{35}-35\sqrt{36}}{1260}\)

= 1 - \(\frac{1}{\sqrt2}\) + \(\frac{1}{\sqrt2}\) - \(\frac{1}{\sqrt3}\) + \(\frac{1}{\sqrt3}\) - \(\frac{1}{\sqrt4}\) + ..............+ \(\frac{1}{\sqrt{35}}\) - \(\frac{1}{\sqrt{36}}\)

= 1 - \(\frac{1}{\sqrt{36}}\) = 1 - \(\frac{1}{6}\) = \(\frac{5}{6}\)

As required 42 S = 42 x \(\frac{5}{6}\) = 35.

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