How do you use the binomial series to expand #1/(2 + x)^2#?

2 Answers

Answer:

#1/(2+x)^2=1/(4+4x+x^2)#

Explanation:

The binomial expansion allows us to take something in the form

#(a+b)^n#

and express it as a sum of terms:

#(a+b)^n=(C_(n,0))a^nb^0+(C_(n,1))a^(n-1)b^1+...+(C_(n,n))a^0b^n#

We can do this for the denominator in the fraction:

#(2+x)^2=1(2)^2(1)+2(2)(x)+1(1)(x^2)=4+4x+x^2#

and so:

#1/(2+x)^2=1/(4+4x+x^2)#

May 9, 2017

Answer:

#1/(2+x)^2=1/4-1/4x+3/16x^2-1/8x^3+5/64x^4-3/64x^5+...............#

Explanation:

#1/(2+x)^2=(2+x)^(-2)=2^(-2)xx(1+x/2)^(-2)#

Now Binomial expansion of #(1+a)^n#

= #1+na+(n(n-1))/(1*2)a^2+(n(n-1)(n-2))/(1*2*3)a^3+(n(n-1)(n-2)(n-3))/(1*2*3*4)a^4+.............#

Hence #(1+x/2)^(-2)#

= #1+(-2)x/2+((-2)(-3))/(1*2)(x/2)^2+((-2)(-3)(-4))/(1*2*3)(x/2)^3+((-2)(-3)(-4)(-5))/(1*2*3*4)(x/2)^4+.............#

= #1-x+3/4x^2-1/2x^3+5/16x^4-3/16x^5+...............#

and #1/(2+x)^2=1/4xx(1-x+3/4x^2-1/2x^3+5/16x^4-3/16x^5+...............)#

= #1/4-1/4x+3/16x^2-1/8x^3+5/64x^4-3/64x^5+...............#