Find the coefficient of #x^7# in the expansion of #(1-x)^(-2)#?

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Answer 1 · 2017-10-15T06:34:42

The coefficient of #x^7# is #8#

The Binomial expansion of #(1+x)^n# is given by

#(1+x)^n=sum_(r=0)^(r=n)C_r^nx^r#,

where #C_r^n=(n!)/(r!(n-r)!)=(n(n-1)(n-2)...(n-r+1))/(1*2*3....*r)#

that is coefficients would be #1,n,(n(n-1))/(1*2),(n(n-1)(n-2))/(1*2*3),.....#

Hence #(1-x)^(-2)=sum_(r=0)^(r=n)C_r^(-2)(-x)^r#

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

and term containing #x^7# woud be

#((-2)(-3)(-4)(-5)(-6)(-7)(-8))/(1*2*3*4*5*6*7)(-x)^7#

= #8x^7#

And the coefficient of #x^7# is #8#