(т-1)л 3 sin 2 [3*sin([(m-1)*pi]/2)]/[(m-1)!] incorrect (т 1)! At least one of the answers above is NOT correct. Find the Maclaurin polynomials of orders n = 0, 1, 2, 3, and 4, and then find the nth...


Last one keeps showing up as incorrect, please help.


(т-1)л<br>3 sin<br>2<br>[3*sin([(m-1)*pi]/2)]/[(m-1)!]<br>incorrect<br>(т<br>1)!<br>At least one of the answers above is NOT correct.<br>Find the Maclaurin polynomials of orders n = 0, 1, 2, 3, and 4, and then find the nth Maclaurin polynomials for the function in sigma notation.<br>Enter the Maclaurin polynomials below for 3x sin(x).<br>Po(x) =<br>P1(x) =<br>P2(x) =<br>3x^(2)<br>P3(x) = 3x^(2)<br>P4(x) = 3x^(2)-(x^(4))/(2)<br>n<br>Pn(x) = 2 (3sin[((m-1)pi)/(2)])/((m-1)!)<br>(Note summation starts at m = 1).<br>m=1<br>

Extracted text: (т-1)л 3 sin 2 [3*sin([(m-1)*pi]/2)]/[(m-1)!] incorrect (т 1)! At least one of the answers above is NOT correct. Find the Maclaurin polynomials of orders n = 0, 1, 2, 3, and 4, and then find the nth Maclaurin polynomials for the function in sigma notation. Enter the Maclaurin polynomials below for 3x sin(x). Po(x) = P1(x) = P2(x) = 3x^(2) P3(x) = 3x^(2) P4(x) = 3x^(2)-(x^(4))/(2) n Pn(x) = 2 (3sin[((m-1)pi)/(2)])/((m-1)!) (Note summation starts at m = 1). m=1

Jun 04, 2022
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