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Algebra Level 2

( 1 + 1 a ) ( 1 + 1 a 2 ) ( 1 + 1 a 4 ) ( 1 + 1 a 8 ) ⋯ ( 1 + 1 a n ) = ? \left(1 + \frac{1}{a}\right) \left(1+\frac{1}{a^2}\right) \left(1 +\frac{1}{a^4}\right) \left( 1 + \frac{1}{a^8}\right) \cdots \left(1+ \frac{1}{a^n}\right) =\, ?

a 2 n a^{2n} a n ( a − 1 ) a^n(a - 1) a 2 n + 1 a n − 1 \frac{a ^{2n} +1}{a^n - 1} a 2 n − 1 a 2 n − 1 ( a − 1 ) \frac{a^{2n} - 1}{a^{2n - 1}(a -1 )}

Vaibhav Singh by vaibhav singh , 18, India

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2 solutions

P = ( 1 + 1 a ) ( 1 + 1 a 2 ) ( 1 + 1 a 4 ) ( 1 + 1 a 8 ) ⋯ ( 1 + 1 a n ) Multiply up and down by 1 − 1 a = ( 1 − 1 a ) ( 1 + 1 a ) ( 1 + 1 a 2 ) ( 1 + 1 a 4 ) ( 1 + 1 a 8 ) ⋯ ( 1 + 1 a n ) 1 − 1 a = ( 1 − 1 a 2 ) ( 1 + 1 a 2 ) ( 1 + 1 a 4 ) ( 1 + 1 a 8 ) ⋯ ( 1 + 1 a n ) 1 − 1 a = ( 1 − 1 a 4 ) ( 1 + 1 a 4 ) ( 1 + 1 a 8 ) ⋯ ( 1 + 1 a n ) 1 − 1 a = ( 1 − 1 a 8 ) ( 1 + 1 a 8 ) ⋯ ( 1 + 1 a n ) 1 − 1 a The process continues = 1 − 1 a 2 n 1 − 1 a = a 2 n − 1 a 2 n − 1 ( a − 1 ) \begin{aligned} P & = \left(1+\frac 1a\right) \left(1+\frac 1{a^2} \right) \left(1+\frac 1{a^4} \right) \left(1+\frac 1{a^8} \right) \cdots \left(1+\frac 1{a^n} \right) & \small \color{#D61F06} \text{Multiply up and down by }1-\frac 1a \\ & = \frac {{\color{#D61F06}\left(1-\frac 1a\right)}{\color{#3D99F6}\left(1+\frac 1a\right)} \left(1+\frac 1{a^2} \right) \left(1+\frac 1{a^4} \right) \left(1+\frac 1{a^8} \right) \cdots \left(1+\frac 1{a^n} \right)}{\color{#D61F06}1-\frac 1a} \\ & = \frac {{\color{#3D99F6}\left(1-\frac 1{a^2} \right) \left(1+\frac 1{a^2} \right)} \left(1+\frac 1{a^4} \right) \left(1+\frac 1{a^8} \right) \cdots \left(1+\frac 1{a^n} \right)}{1-\frac 1a} \\ & = \frac {{\color{#3D99F6}\left(1-\frac 1{a^4} \right) \left(1+\frac 1{a^4} \right)} \left(1+\frac 1{a^8} \right) \cdots \left(1+\frac 1{a^n} \right)}{1-\frac 1a} \\ & = \frac {{\color{#3D99F6}\left(1-\frac 1{a^8} \right) \left(1+\frac 1{a^8} \right)} \cdots \left(1+\frac 1{a^n} \right)}{1-\frac 1a} & \small \color{#3D99F6} \text{The process continues} \\ & = \frac {1-\frac 1{a^{2n}}}{1-\frac 1a} \\ & = \boxed{\dfrac {a^{2n}-1}{a^{2n-1}(a-1)}} \end{aligned}

1 Helpful 0 Interesting 0 Brilliant 0 Confused
Vaibhav Singh
Jul 22, 2018

Let P = (1 + 1 a \frac{1}{a} )(1 + 1 a 2 \frac{1}{a^2} )..... Now multiply by (1 - 1 a \frac{1}{a} ) both side then you will see a pattern which you have studied i.e. (x -y)(x +y) = x^2 - y^2 this means that (1 - 1 a \frac{1}{a} )P = (1 - 1 a \frac{1}{a} )(1 + 1 a \frac{1}{a} )(1 + 1 a 2 \frac{1}{a^2} )...(1+` 1 a 4 \frac{1}{a^4} ) you can clearly see that (1 - 1 a \frac{1}{a} )(1+ 1 a \frac{1}{a} ) becomes (1 - 1 a 2 \frac{1}{a^2} ) similarly it is multiplied by (1 + 1 a 2 \frac{1}{a^2} ) then it becomes (1- 1 a 4 \frac{1}{a^4} ) this pattern goes on and at last it becomes (1 - 1 a \frac{1}{a} )P = (1- 1 a n \frac{1}{a^n} )(1+ 1 a n \frac{1}{a^n} ) which can be simplified and its become P* = 1 − 1 / a 2 n 1 − 1 / a \frac{1-1/a^2n}{1 - 1/a} * - As i am new to brilliant i don't know how to write a to the power 2n hence by mistake it was written as a to the power 2 multiplied n .CORRECT IT Now this can simplified and you will get the answer a ( a 2 n − 1 ) a 2 n ( a − 1 \frac{a(a^2n - 1)}{a^2n(a - 1}

0 Helpful 0 Interesting 0 Brilliant 0 Confused

In LaTeX, use braces { } around the powers if it's more than one character. Therefore a^{2n} becomes a 2 n a^{2n}

Also note that it's better to put whole equations into LaTeX rather than small pieces. LaTeX is designed to format your equations for you, and while it may not look perfect at first, you'll learn small bits of LaTeX as time goes on to make it look better. If you want to write mathematics documents in the future, this will be a useful skill outside of brilliant.org

Brian Moehring - 2 years, 10 months ago

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