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Answer 1 · 2018-01-19T12:48:02

$(4-3x^2)^(-1/2)=1/2+(3x^2)/16+(27x^4)/256+(135x^6)/2048$

$(a+b)^n=a^n(1+b/a)^n=a^n(1+n(b/a)+(n(n-1)(b/a)^2)/(2!)+(n(n-1)(n-2)(b/a)^3)/(3!)+cdots)$

For $(4-3x^2)^(-1/2)$

We have $4^(-1/2)(1-(3x^2)/4)^(-1/2)=(1-(3x^2)/4)^(-1/2)/2$

$(1-(3x^2)/4)^(-1/2)=1+(-1/2)((-3x^2)/4)+((-1/2)(-3/2)((-3x^2)/4)^2)/(2!)+((-1/2)(-3/2)(-5/2)((-3x^2)/4)^3)/(3!)$

$color(white)((1-(3x^2)/4)^(-1/2))=1+(3x^2)/8+((3/4)(9x^4)/16)/2+((-15/8)(-27x^6)/64)/6$

$color(white)((1-(3x^2)/4)^(-1/2))=1+(3x^2)/8+((27x^4)/64)/2+((405x^6)/512)/6$

$color(white)((1-(3x^2)/4)^(-1/2))=1+(3x^2)/8+(27x^4)/128+(405x^6)/3072$

$color(white)((1-(3x^2)/4)^(-1/2))=1+(3x^2)/8+(27x^4)/128+(135x^6)/1024$

$1/2(1-(3x^2)/4)^(-1/2)=(1+(3x^2)/8+(27x^4)/128+(135x^6)/1024)/2=1/2+(3x^2)/16+(27x^4)/256+(135x^6)/2048$