Question-145918

Question Number 145918 by mathdanisur last updated on 09/Jul/21

Answered by Olaf_Thorendsen last updated on 09/Jul/21

l = 3 + \sqrt{3^{2} + \sqrt{3^{4} + \sqrt{3^{8} + \sqrt{\dots}}}}

{(l - 3)}^{2} = 3^{2} + \sqrt{3^{4} + \sqrt{3^{8} + \sqrt{\dots}}}

{(l - 3)}^{2} = 3^{2} + \sqrt{3^{4} + 3^{2} \sqrt{3^{4} + \sqrt{\dots}}}

{(l - 3)}^{2} = 3^{2} + 3 \sqrt{3^{2} + \sqrt{3^{4} + \sqrt{\dots}}}

\frac{{(l - 3)}^{2}}{3} = 3 + \sqrt{3^{2} + \sqrt{3^{4} + \sqrt{\dots}}} = l

l^{2} - 9 l + 9 = 0

l = \frac{9 \pm \sqrt{81 - 4 \times 1 \times 9}}{2}

l = \frac{9 \pm 3 \sqrt{5}}{2}

\frac{9 - 3 \sqrt{5}}{2} \approx 1, 146 : impossible because l > 3

Finally l = \frac{9 + 3 \sqrt{5}}{2} \approx 7, 854

Commented by Mrsof last updated on 09/Jul/21

s i r c a n y o u h e l p m e b y r e s i d e o

Commented by Snail last updated on 09/Jul/21

y o u a r e t h a t i n s t a g r a m g u y w i t h t h e i d g e r c e k b o s

Commented by mathdanisur last updated on 09/Jul/21

T h a n k y o u S e r, c o o l