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Mathematics 16 Online
OpenStudy (anonymous):

Factorise the following: 2^(n+1)+3(2^n)+2^(n-1)

OpenStudy (lalaly):

\[2^{n+1}=2(2^n)\]\[2^{n-1}=2^{-1}2^n=\frac{1}{2}(2^n)\]now u have\[2(2^n)+3(2^n)+\frac{1}{2}(2^n)\]now factor out 2^n\[=2^n(2+3+\frac{1}{2})=\frac{11}{2}(2^n)\]

OpenStudy (anonymous):

\[ \frac {11} 2 2^n= 11\, \,( 2^{n-1}) \]

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