Find the Maclaurin series for using the definition of a Maclaurin series. [Suppose it has a power series expansion. Do not show that Rn(x) tends to 0.] Find the related radius of convergence as well.&nbsp;f(x)=(1−x)−2

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Find the Maclaurin series for using the definition of a Maclaurin series. [Suppose it has a power series expansion. Do not show that Rn(x) tends to 0.] Find the related radius of convergence as well.&amp;nbsp;f(x)=(1−x)−2

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Find a few derivatives and determine the values of each one at&amp;nbsp;a=0. Remember that there is a (-1) that comes from the derivative of&amp;nbsp;(1−x)f(x)=(1−x)−2f′(x)=−2(1−x)−3(−1)=2(1−x)−3f″(x)=(−3)2(1−x)−4(−1)=(3)(2)(1−x)−4f‴(x)=(−4)(3)2(1−x)−5(−1)=(4)(3)(2)(1−x)−5f4(x)=(−5)(4)(3)(2)(1−x)−6(−1)=(5)(4)(3)(2)(1−x)−5fn(x)=(n+1)!(1−x)−(n+2)Plug everything into Maclaurin general form:f(x)=f(a)+f′(a)1!x+f″(a)2!x2+f‴(a)3!x3+...f(x)=1+2x+3!2!x2+4!3!x3+5!4!x4+...=1+2x+3x2+4x3+5x4+...Find the pattern of the numbers to write in summation formf(x)=∑(n=0)∞(n+1)xn&amp;nbsp;Use the Ratio Test&amp;nbsp;|an+1an|=|((n+1)+1)xn+1(n+1)xn|=|(n+2)xn+1(n+1)xn|=(n+2)|x|n+1limn→∞(n+2)|x|(n+1)⋅1n1n=limn→∞(1+2n)|x|1+1n=(1+0)|x|(1+0)=|x|This converges when|x|&amp;lt;1The radius of convergence is half the width of this intervalR=1−(−1)2=1Resultf(x)=∑n=0∞(n+1)xn,R=1

Find the Maclaurin series for using the definition of a Maclaurin series. [Assume that has a power series expansion.Do not show that Rn(x) tends to 0.] Also find the associated radius of convergence. f(x)=(1-x)^{-2}

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