Tom K. answered 05/10/20
Knowledgeable and Friendly Math and Statistics Tutor
The nth derivative of 1/(6+x)^3 will be (-1)^n n+2!/2! (6+x)^-(3+n)
Thus, the Taylor series (recall the division by n!) is
n=∞
∑ (-1)^n(n+2)(n+1)/(2*6^(n+3))*x^n
n=0
Alyssa H.
asked 05/10/20Use differentiation and/or integration to express the following function as a power series (centered at x=0).
f(x)=1/((6+x)^3)
∞
f(x)=∑ _________
n=0
Tom K. answered 05/10/20
Knowledgeable and Friendly Math and Statistics Tutor
The nth derivative of 1/(6+x)^3 will be (-1)^n n+2!/2! (6+x)^-(3+n)
Thus, the Taylor series (recall the division by n!) is
n=∞
∑ (-1)^n(n+2)(n+1)/(2*6^(n+3))*x^n
n=0
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