Find the 2

Find the term independent of x in the expansion of (4x3+1/2x)8


x =  7


P=(4x3+12x)8 X=(4x3 1/(2x) 1/(2x) 1/(2x))2 X=(4 1/2 1/2 1/2)2=14=0.25  C2(8)=(82)=8!2!(82)!=8721=28  x=(82) X=28 0.25=7P=(4x^3+\dfrac{ 1 }{ 2x } )^8 \ \\ X=(4x^3 \cdot \ 1/(2x) \cdot \ 1/(2x) \cdot \ 1/(2x))^{ 2 } \ \\ X=(4 \cdot \ 1/2 \cdot \ 1/2 \cdot \ 1/2)^{ 2 }=\dfrac{ 1 }{ 4 }=0.25 \ \\ \ \\ C_{{ 2}}(8)=\dbinom{ 8}{ 2}=\dfrac{ 8! }{ 2!(8-2)!}=\dfrac{ 8 \cdot 7 } { 2 \cdot 1 }=28 \ \\ \ \\ x={ { 8 } \choose 2 } \cdot \ X=28 \cdot \ 0.25=7

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