How do you find the maclaurin series expansion of #f(x) = 1/((1+x)^2)#?

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Answer 1 · 2017-01-11T16:14:19

The answer is #==1-2x+3x^2-4x^3+5x^4+....#

The Maclaurin series is

#f(x)=f(0)+(f'(0))/(1!)x+(f''(0))/(2!)x^2+(f'''(0))/(3!)x^3+(f^(iv)(0))/(4!)x^4+..#

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

#f(0)=1#

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

#f'(0)=-2#

#f''(x)=6/(1+x)^4#

#f''(0)=6#

#f'''(x)=-24/(1+x)^5#

#f'''(0)=-24#

#f^(iv)(x)=120/(1+x)^6#

#=1+(-2)/1x+(6)/(2!)x^2+(-24)/(3!)x^3+(+120)/(4!)x^4+....#

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