Ramanujan's nested radical is great!

Algebra Level 4

Let S = x = 0 62 1 + x 1 + ( x + 1 ) 1 + ( x + 2 ) 1 + S= \displaystyle\sum_{x=0}^{62}\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+2)\sqrt{1+\cdots}}}}

Hence, find S \lfloor S\rfloor

Hints: 3 = 1 + 2 1 + 3 1 + 4 1 + 3=\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+\cdots}}}}

2 + 1 = 1 + 2 1 + ( 2 + 1 ) 1 + ( 2 + 2 ) 1 + 2+1=\sqrt{1+2\sqrt{1+(2+1)\sqrt{1+(2+2)\sqrt{1+\cdots}}}}


The answer is 2016.

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1 solution

Tommy Li
May 26, 2016

x + 1 = 1 + x 1 + ( x + 1 ) 1 + ( x + 2 ) 1 + . . . x+1=\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+2)\sqrt{1+...}}}} (by Ramanujan)

i = 0 62 1 + x 1 + ( x + 1 ) 1 + ( x + 2 ) 1 + . . . = 1 + 2 + 3 + . . . + 63 = ( 63 + 1 ) ( 63 ) 2 = 2016 \sum_{i=0}^{62}\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+2)\sqrt{1+...}}}}=1+2+3+...+63=\frac{(63+1)(63)}{2}=2016

2016 = 2016 \lfloor 2016\rfloor =2016

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