An Exponentially Little Christmas Present

Algebra Level 3

For n ≠ 0 n \ne 0 , given that x = 1 2 ( 3 1 / n − 3 − 1 / n ) x = \dfrac12 \left( 3^{1/n} - 3^{-1/n} \right) , find the value of ( x + 1 + x 2 ) n \left(x + \sqrt{1+x^2} \right)^n .


The answer is 3.

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Ponhvoan Srey by Ponhvoan Srey , 21, Singapore

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2 solutions

Jonathan Alvaro
Dec 26, 2015

Plug in n = 1 and you get x=4/3. Plug both n and x into the second equation and you get the answer

2 Helpful 0 Interesting 0 Brilliant 0 Confused

Nice! Short and sweet!

Ponhvoan Srey - 5 years, 5 months ago
Ponhvoan Srey
Dec 23, 2015

x = 1 2 ( 3 1 n − 3 − 1 n ) x 2 = 1 4 ( 3 2 n + 3 − 2 n − 2 ) = 1 4 ( 3 2 n + 3 − 2 n ) − 1 2 x 2 + 1 = 1 4 ( 3 2 n + 3 − 2 n ) + 1 2 = ( 1 2 ( 3 1 n + 3 − 1 n ) ) 2 x 2 + 1 = 1 2 ( 3 1 n + 3 − 1 n ) x + x 2 + 1 = 1 2 ( 3 1 n − 3 − 1 n ) + 1 2 ( 3 1 n + 3 − 1 n ) x + x 2 + 1 = 3 1 n ⇒ ( x + x 2 + 1 ) n = 3 \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad x=\frac { 1 }{ 2 } \left( { 3 }^{ \frac { 1 }{ n } }-{ 3 }^{ -\frac { 1 }{ n } } \right) \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad { x }^{ 2 }=\frac { 1 }{ 4 } \left( { 3 }^{ \frac { 2 }{ n } }{ +3 }^{ -\frac { 2 }{ n } }-2 \right) =\frac { 1 }{ 4 } \left( { 3 }^{ \frac { 2 }{ n } }{ +3 }^{ -\frac { 2 }{ n } } \right) -\frac { 1 }{ 2 } \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad { x }^{ 2 }+1=\frac { 1 }{ 4 } \left( { 3 }^{ \frac { 2 }{ n } }{ +3 }^{ -\frac { 2 }{ n } } \right) +\frac { 1 }{ 2 } ={ \left( \frac { 1 }{ 2 } \left( { 3 }^{ \frac { 1 }{ n } }+{ 3 }^{ -\frac { 1 }{ n } } \right) \right) }^{ 2 }\\ \quad \quad \quad \quad \quad \quad \quad \sqrt { { x }^{ 2 }+1 } =\quad \frac { 1 }{ 2 } \left( { 3 }^{ \frac { 1 }{ n } }+{ 3 }^{ -\frac { 1 }{ n } } \right) \\ \quad \quad \quad \quad \quad x+\sqrt { { x }^{ 2 }+1 } =\frac { 1 }{ 2 } \left( { 3 }^{ \frac { 1 }{ n } }-{ 3 }^{ -\frac { 1 }{ n } } \right) +\frac { 1 }{ 2 } \left( { 3 }^{ \frac { 1 }{ n } }+{ 3 }^{ -\frac { 1 }{ n } } \right) \quad \\ \quad \quad \quad \quad \quad x+\sqrt { { x }^{ 2 }+1 } ={ 3 }^{ \frac { 1 }{ n } }\\ \Rightarrow \boxed { { \left( x+\sqrt { { x }^{ 2 }+1 } \right) }^{ n }=3 }

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