Alternate method to solve $\sqrt{11\sqrt{11\sqrt{11...4\, \text{times}}}}$
Question :
What is the value of$$\sqrt{11\sqrt{11\sqrt{11...4\,\text{times}}}}$$
I did it by solving square root one by one.$$\sqrt{11\sqrt{11\sqrt{11\times11^\frac{1}{2}}}}$$$$\sqrt{11\sqrt{11\sqrt{11^\frac{3}{2}}}}$$$$\sqrt{11\sqrt{11\times{11^\frac{3}{4}}}}$$$$\sqrt{11\sqrt{11^\frac{7}{4}}}$$$$\sqrt{11\times{11^\frac{7}{8}}}$$$$\sqrt{11^\frac{15}{8}}$$$$11^\frac{15}{16}$$
Is there any other way to solve this?
I don't want the complete solution, just tell me the approach.
$\endgroup$ 112 Answers
$\begingroup$If you solve the square roots one by one, but from outer one to inner one, you see it is the sum of geometric sequence.
$\endgroup$ $\begingroup$There is a faster way to solve it, actually:
$\sqrt{11·\sqrt{11·\sqrt{11·\sqrt{11·\color{blue}{11}}}}} = \sqrt{11·\sqrt{11·\sqrt{11·11}}} = \sqrt{11·\sqrt{11·11}} = \sqrt{11·11} = 11$
So clearly the extra factor is just $\sqrt{\sqrt{\sqrt{\sqrt{\color{blue}{11}}}}}$, since square-root commutes with multiplication on non-negative reals (which implies that in any expression comprising just multiplication and square-roots the number of square-roots in effect is just the number of 'radical lines' above it).
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