Genius of Srinivasa Ramanujan

  1. In December 1914, Ramanujan was asked by his friend P.C. Mahalanobis to solve a puzzle that appeared in Strand magazine as “Puzzles at a Village Inn”. The puzzle stated that n houses on one side of the street are numbered sequentially starting from 1. The sum of the house numbers on the left of a particular house having the number m, equals that of the houses on the right of this particular house. It is given that n lies between 50 and 500 and one has to determine the values of m and n. Ramanujan immediately rattled out a continued fraction generating all possible values of m without having any restriction on the values of n. List the first five values of m and n.
  2. Ramanujan had posed the following problem in a journal: \sqrt{1+2\sqrt{1+3\sqrt{\ldots}}}=x, find x. Without receiving an answer from the readers, after three months he gave answer as 3. This he could say because he had an earlier general result stating 1+x=\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+2)\sqrt{1+\ldots}}}} is true for all x. Prove this result, then x=2 will give the answer to Ramanujan’s problem.

Try try until you succeed!!

Nalin Pithwa.

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