Question

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Prove :-

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

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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,

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

Hence, it is proved.

Answer 2

hello there

Step-by-step explanation:

Question:

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

Solution:

Taking LHS :

we know that :

$= > 3 = \sqrt{9}$

We can further write the expression as :

$3 = \sqrt{9} = \sqrt{1 + 8}$

Again, we can further do modification in the expression as :

$3 = \sqrt{1 + 8} = \sqrt{1 + (2 \times 4)}$

Also , √16 = 4

So, replacing the value of 4 with 16

Which is also equal to :

$3 = \sqrt{1 + 2 \sqrt{16} }$

Now , 16 can be written as = 1 +15

So , replacing 16 too from the expression:

$3 = \sqrt{1 + 2 \sqrt{1 + 15} }$

Now , in this expression we can write 15 as = 3×5

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

Now , similarly as we have done above applying the same technique:

As we know √25 = 5

So, putting the value instead of 5

we will get :

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

Now , 25 = 1 + 24 ;

Placing the value in the expression:

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

Also , 24 = 6 ×4

Therefore, we will get :

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

Now , Again √36 = 6

So, replacing its value from the expression:

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

Here ,

[ 36 can be written as = 1+35 ]

If , we continue to replace values again and again : we will get an expression as :

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

Which is equal to R.H.S

[ Hence proved ]

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Hope it helps

Thank you!