Simplify #(4^n - 4^(n-2))/4^(n+1)# ?

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Answer 1 · 2018-05-29T12:44:48

#=>15/64#

#(4^n - 4^(n-2))/(4^(n+1))#

We can factor out #4^n# in both the numerator and denominator

#=(4^n(4^0-4^(-2)))/(4^n * 4^1)#

This factor will cancel now

#={cancel(4^n)(1-1/16)} / {cancel(4^n)*4}#

We can simplify down to a single fraction

#=(15/16)/4#

#=15/64#