1.  The series is ∑_n=0^∞a_nxⁿ,where a_n = (-1)ⁿ(n+1)(n+2)/2

1. For f(x) = 1/(1 + x)³ we compute some derivatives and seek a pattern. f'(x)= -3/(1 + x)⁴ f”(x)= 3⋅4/(1 + x)⁵ f”'(x)= -3⋅4⋅5/(1 + x)⁶ and so on. Evaluating the function and its derivatives at x = 0 we find f(0)= 1 f'(0)= -3 f”(0)= 3⋅4 f”'(0)= -3⋅4⋅5 and so on. Then the coefficients of the Taylor series, f(n)(0)/n!, are f(0) = 1 f'(0)/1! = -3 f”(0)/2! = 3⋅4/2! = 3⋅4/2 f”'(0)/3! = -3⋅4⋅5/3! = – 4⋅5/2 We need a few more terms to find the pattern. f(⁴)(0)/4! = 3⋅4⋅5⋅6/4! = 5⋅6/2 f(⁵)(0)/5! = -3⋅4⋅5⋅6⋅7/5! = -6⋅7/2 Now we see it: f(ⁿ)(0)/n! = (-1)ⁿ(n+1)(n+2)/2

1. The expansion is f(x)= 1 -2(x-1) -3(x-1)² – (x-1)³.

1. We compute the derivatives. Only a few are nonzero, so we need to expend much energy looking for the pattern. f(1)= 1 f'(1)= -2 f”(1)= -6 f”'(x)= -6 f(ⁿ)(x) = 0 for all n > 3. Then the Taylor series is f(x) = 1 – (2/1!)(x-1) – (6/2!)(x-1)² – (6/3!)(x – 1)³ = 1 -2(x-1) -3(x-1)² – (x-1)³

1. The series is x³ + x⁴ + x⁵/2! + x⁶/3! + x⁷/4! + … + xⁿ⁺³/n! + …

1. Recall the Taylor series for e^x= 1 + x + x²/2! + x³/3! + x⁴/4! + … Multiplying by x³ raises all the exponents by 3: x³e^x= x³ + x⁴ + x⁵/2! + x⁶/3! + x⁷/4! + … + xⁿ⁺³/n! + …

1. sin²(x)= -x² + 2³x⁴/4! – 2⁵x⁶/6! + …

1. Recall sin²(x) = (1 – cos(2x))/2. The Taylor series for cos(2x) is cos(2x)= 1 – (2x)²/2! + (2x)⁴/4! – (2x)⁶/6! + ….So 1 – cos(2x) has series expansion 1 – cos(2x)= -2²x²/2! + 2⁴x⁴/4! – 2⁶x⁶/6! + …and we find sin²(x)= -x² + 2³x⁴/4! – 2⁵x⁶/6! + …

1. The limit is -1/6.

1. Substitute in the Taylor series for sin(x), obtaining sin(x) – x = (x -x³/3! + x⁵/5! – …) – x= -x³/3! + x⁵/5! – … and so (sin(x) – x)/x³ = -1/3! + x²/5! – … where all the omitted terms have powers of x higher than 2, specifically, x⁴, x⁶, x⁸ and so on. As x → 0, only the -1/3! remains.