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TOT 302 1991 Autumn O J3 nested fractions sum

Source:

June 9, 2024
algebraSum

Problem Statement

Prove that 12+13+14+1...+19991+11+11+13+14+1...+19991=1\dfrac{1}{2+\dfrac{1}{3+\dfrac{1}{4+\dfrac{1}{...+\dfrac{1}{9991}}}}}+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{3+\dfrac{1}{4+\dfrac{1}{...+\dfrac{1}{9991}}}}}}=1
This means 1/(2+(1/(3+(1/(4+(...+1/1991))))))+1/(1+(1/(1+(1/(3+(1/(4+(...+1/1991...))))))))=1.)1/(2+ (1/(3+ (1/(4+(...+1/1991)))))) +1/(1 + (1/(1 + (1/(3 + (1/(4 + (...+ 1/1991...)))))))) = 1.)
(G. Galperin, Moscow-Tel Aviv)