Question

A challenge for all brainlics! ☺☺

Prove :-

$3 = \sqrt{1 + 2 \sqrt{1 + 3 \sqrt{1 + 4 \sqrt{1 + ...} } } }$


No Spam .

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Answer 1

Answer:

$3 = \sqrt{1 + 2 \sqrt{1 + 3 \sqrt{1 + 4 \sqrt{1 + ...} } } }$

Step-by-step explanation:

We can write,

$\begin{lgathered}\sf \because 3 = \sqrt{9} . \\ \\ \sf = \sqrt{1 + 8} . \\ \\ \sf = \sqrt{1 + 2 \times 4} . \\ \\ \sf = \sqrt{1 + 2 \sqrt{16} } . \\ \\ \sf \sf = \sqrt{1 + 2 \sqrt{1 + 15} } . \\ \\ \sf = \sqrt{1 + 2 \sqrt{1 + 3 \times 5} } . \\ \\ \sf= \sqrt{1 + 2\sqrt{1 + 3 \sqrt{25} } } . \\ \\ \sf = \sqrt{1 + 2 \sqrt{1 + 3 \sqrt{1 + 24} } } . \\ \\ \sf = \sqrt{1 + 2 \sqrt{1 + 3 \sqrt{1 + 4 \times 6} } } . \\ \\ \sf = \sqrt{1 + 2 \sqrt{1 + 3 \sqrt{1 + 4 \sqrt{36} } } } . \\ \\ \pink{ \boxed{ \orange{ \sf = \sqrt{1 + 2 \sqrt{1 + 3 \sqrt{1 + 4 \sqrt{1 + ......} } } } .}}}\end{lgathered}$

Hence, it is proved.

Answer 2

ANSWER:-

➣

• Given:-

$3 = \sqrt{1 + 2 \sqrt{1 + 3 \sqrt{1 + 4 \sqrt{1 + ...} } } }$

The following can be written as:-

$\longrightarrow{3 = \sqrt{9}}$

We have:-

$= \sqrt{1 + 8}$

$= \sqrt{1 + 2 \times 4}$

$= \sqrt{1 + 2 \sqrt{16} }$

$= \sqrt{1 + 2 \sqrt{1 + 15} }$

$= \sqrt{1 + 2 \sqrt{1 + 3 \times 5} }$

$= \sqrt{1 + 2\sqrt{1 + 3 \sqrt{25} } }$

$= \sqrt{1 + 2 \sqrt{1 + 3 \sqrt{1 + 24} } }$

$= \sqrt{1 + 2 \sqrt{1 + 3 \sqrt{1 + 4 \times 6} } }$

$= \sqrt{1 + 2 \sqrt{1 + 3 \sqrt{1 + 4 \sqrt{36} } } }$

${ \boxed{\bf{= \sqrt{1 + 2 \sqrt{1 + 3 \sqrt{1 + 4 \sqrt{1 + ......} } } } }}}$

Hence Proved.✅