q4 6.2 Is the solution correct?Question:Use the fact that(1+2)?= E(-1)" (n + 1) 2" (converges for -1 < x < 1) to find a series that converges to(1+2)(for -1 < z < 1).れ=0"Solution":Line 1: Notice that if f (x) =(1+2) then f' (x)(1+2)Line 2: Therefore, to find a series forwe could differentiate the series for(1+2)3(1+2)?Line 3: To find the derivative of a convergent series, we can differentiate term-by-term (i.e. differentiate one term at a time, as if we were differentiating a polynomial).Line 4: Let's expand the given series:Line 5: 1+2)6x5+...= 1- 2x + 3x2 – 4x3 + 5x4Line 6: Now differentiate both sides:Line 7:= -2 + 3. 2x – 4. 3x2 + 5 . 4x3 – 6- 5x4+...(1+2)3Line 8: Write using sigma notation:Line 9: -2E(-1)" (n + 1) - n - 2"n=0Line 10: The radius of convergence does not change when differentiating, so E (-1)" (n + 1) · n· x" is a power series that converges to(1+a)'for -1 < a < 1.n=0Is the above "solution" correct? If not, in which line is the first error? [ Select ]What is the correct final answer to the question? (Choose from the options below) [ Select]DOOption I: E (-1)" (n + 1) · n · x"n=0Option II: E (-1)" (n +1) - n · r"-1n=1Option III:E(-1)" (n - 1) · n xn+1n=0Option IV: E (-1)"+1 (n + 1) · n · x"-1n=1

Answered: Image /qna-images/answer/6f63a28b-8f56-4fb6-b460-bf6c944db07e.jpg

Is the solution correct? Question: 8, E(-1)" (n + 1) x" (converges for -1 < x < 1) to find a series that converges to (for -1 < x < 1). Use the fact that %3D (1+2)* (1+2) "Solution": Line 1: Notice that if f (x) = y then f' (x) (1+x)* (1+2) Line 2: Therefore, to find a series for -2 we could differentiate the series for (1+2)3 (1+a) Line 3: To find the derivative of a convergent series, we can differentiate term-by-term (i.e. differentiate one term at a time, as if we were differentiating a polynomial). Line 4: Let's expand the given series: = 1- 2x + 3x2 – 4x3 + 5x4 (1+z) 6x5 +... Line 5: Line 6: Now differentiate both sides: Line 7: (1+2) = -2 + 3. 2 – 4. 3x2 + 5 . 4x3 – 6· 5x4 +... Line 8: Write using sigma notation: Line 9: = E (-1)" (n + 1) · n. x" (1+z) n=0 Line 10: The radius of convergence does not change when differentiating, so E (-1)" (n + 1) · n. x" is a power series that converges to -2 for -1