Radicals are Fun!!!

Algebra Level 4

x n − 1 ( x n − 2 ( x n − 3 ⋯ x n − 2 n − 1 n \sqrt[n]{x^{n-1} \big ( \sqrt[n-1]{ x^{n-2} \big ( \sqrt[n-2]{x^{n-3} \cdots \sqrt x } }}

Consider the nested expression above. When n = 12345678987654321 n=12345678987654321 , the expression can be simplified to x 1 − 1 / a ! x^{1- 1/a!} .

Compute a 100 \dfrac a{100} .

For instance, for n = 5 n=5 , the expression is in the form x 4 ( x 3 ( x 2 ( x 3 4 5 \sqrt [ 5 ]{ { x }^{ 4 }(\sqrt [ 4 ]{ { x }^{ 3 }(\sqrt [ 3 ]{ { x }^{ 2 }(\sqrt { x } } } } .

This problem is original.


The answer is 123456789876543.21.

Aditya Dhawan by Aditya Dhawan , 20, India

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2 solutions

Rishabh Jain
Feb 17, 2016

Start from rightmost end (i.e x \sqrt x ). x n − 1 ( ⋯ x 4 ( x 3 ( x 2 + 1 2 3 4 5 n − 1 n \sqrt[n]{ x^{n-1}(\sqrt[n-1]{\cdots\sqrt [ 5 ]{ { x }^{ 4 }(\sqrt [ 4 ]{ { x }^{ 3 }(\sqrt [ 3 ]{ { x }^{ 2+\color{#D61F06}{\frac{1}{2}}}} }}} } = x n − 1 ( ⋯ x 4 ( x 3 + 5 6 4 5 n − 1 n =\sqrt[n]{ x^{n-1}(\sqrt[n-1]{\cdots\sqrt [ 5 ]{ { x }^{ 4 }(\sqrt [ 4 ]{ { x }^{ 3+\color{#D61F06}{\frac{5}{6} }}} }}} = x n − 1 ( ⋯ x 4 + 23 24 5 n − 1 n =\sqrt[n]{ x^{n-1}(\sqrt[n-1]{\cdots\sqrt [ 5 ]{ { x }^{ 4+\color{#D61F06}{\frac{23}{24} }}} }} ⋯ \cdots = x ( n − 1 ) + ( n − 1 ) ! − 1 ( n − 1 ) ! n =\sqrt[n]{x^{(n-1)+\color{#D61F06}{\frac{(n-1)!-1}{(n-1)!}}}} = x 1 − 1 n ! \large =x^{\color{#D61F06}{^{1-\frac{1}{n!}}}} Whatever we put for 'n', that only we would get for 'a' also . Hence for n=123...321 we'll get a=123...321. 12345668987654321 100 = 123456789876543.21 \huge \frac{12345668987654321}{100}=\boxed{\color{#007fff}{123456789876543.21}}

3 Helpful 0 Interesting 0 Brilliant 0 Confused
Aditya Dhawan
Feb 16, 2016

L e t x n − 1 ( x n − 2 ( x n − 3 . . . . . . . . . . . . x n − 2 n − 1 n = y T h e n , y n ! = x ( n − 1 ) [ ( n − 1 ) ! ] + ( n − 2 ) [ ( n − 2 ) ! ] . . . . . . . + 2 ( 2 ! ) + 1 ( 1 ! ) ( S u c c e s s i v e l y r a i s i n g b o t h s i d e s b y n , n − 1...2 ) ⇒ y = ( x ) ∑ i = 1 n − 1 i ( i ! ) n ! ( 1 ) N o w , ∑ i = 1 n − 1 i ( i ! ) = ∑ ( i + 1 − 1 ) ( i ! ) = ∑ i ! ( i + 1 ) − i ! = ∑ ( i + 1 ) ! − i ! = 2 ! − 1 ! + 3 ! − 2 ! . . . + ( n − 1 ) ! − ( n − 2 ) ! + n ! − ( n − 1 ) ! = n ! − 1 ( T e l e s c o p i n g S e r i e s ) T h u s y = x n ! − 1 n ! = x 1 − 1 n ! N o w n = 12345678987654321 T h u s t h e a n s w e r s h o u l d b e = 12345678987654321 100 = 123456789876543.21 Let\quad \sqrt [ n ]{ { x }^{ n-1 }(\sqrt [ n-1 ]{ { x }^{ n-2 }(\sqrt [ n-2 ]{ { x }^{ n-3 }............\sqrt { x } } } } \quad =\quad y\\ \\ Then,\quad \\ { y }^{ n! }=\quad { x }^{ (n-1)[(n-1)!]+(n-2)[(n-2)!].......+2(2!)+1(1!) }\quad (\quad Successively\quad raising\quad both\quad sides\quad by\quad n,n-1...2)\\ \\ \Rightarrow y=({ x) }^{ \frac { \sum _{ i=1 }^{ n-1 }{ i(i!) } }{ n! } }(1)\\ Now,\\ \sum _{ i=1 }^{ n-1 }{ i(i!) } =\quad \sum { (i+1-1)(i!) } =\sum { i!(i+1)-i!=\sum { (i+1)!-i!=2!-1!+3!-2!...+(n-1)!-(n-2)!+n!-(n-1)!=n!-1\quad } } (Telescoping\quad Series)\\ Thus\quad y=\quad { x }^{ \frac { n!-1 }{ n! } }=\quad { x }^{ 1-\frac { 1 }{ n! } }\\ Now\quad n=12345678987654321\\ Thus\quad the\quad answer\quad should\quad be=\frac { 12345678987654321 }{ 100 } =123456789876543.21

2 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

When you want to show that a pattern exists (without explicitly stating it), it is best to have each of the terms be in the exact form of the pattern. IE, instead of 4, write it as 2 × 2 ! 2 \times 2 ! . Similarly, instead of 1, write it as 1 × 1 ! 1 \times 1! .

Note: You should only Latex up the equations. There is no need to Latex up the text, which makes it harder to read.

Calvin Lin Staff - 5 years, 3 months ago

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