How do you use the binomial series to expand #(1+x)^-3#?

1 Answer
Oct 29, 2017

Answer:

#(1+x)^-3 = ( 1 + 3x+3x^2+x^3 )^-1#

Explanation:

#(1+x)^-3 = 1/(1+x)^3#

Binomial theorem:

#(a + b)^n =(nC_0) a^nb^0+ (nC_1) a^(n-1)b^1 +(nC_2) a^(n-2)b^2 +...... (nC_n)b^n#

Here # (a=1 ; b=x , n= 3) :. 1^n=1 , x^0=1#

We know #nC_r= (n!)/(r!(n-r)!) :. 3C_0=1 , 3C_1=3 , 3C_2=3 ,3C_3=1#

#:. (1+x)^3= 1^3 + 3*1^2*x^1+3*1^1*x^2+1*1^0x^3 # or

#:. (1+x)^3= 1 + 3x+3x^2+x^3 :.#

#(1+x)^-3 = 1/(1+x)^3 = 1/( 1 + 3x+3x^2+x^3 )# or

#(1+x)^-3 = ( 1 + 3x+3x^2+x^3 )^-1# [Ans]

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