3x2+x+1x-14=Ax-1+Bx-12+Cx-13+Dx-14 then A+B+C=

3x2+x+1x-14=Ax-1+Bx-12+Cx-13+Dx-14x-1=y ⇒ x=y+13(y+1)2+y+1+1y4=3y2+7y+5y4 =3y2+7y3+5y4 =3(x-1)2+7(x-1)3+5(x-1)4A=0, B=3, C=7, D=5A+B-C+D=0+3-7+5=1

3x2+x+1x-14=Ax-1+Bx-12+Cx-13+Dx-14 then A+B+C=

3x2+x+1x-14=Ax-1+Bx-12+Cx-13+Dx-14x-1=y ⇒ x=y+13(y+1)2+y+1+1y4=3y2+7y+5y4 =3y2+7y3+5y4 =3(x-1)2+7(x-1)3+5(x-1)4A=0, B=3, C=7, D=5A+B-C+D=0+3-7+5=1

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$\frac{3 x^{2} + x + 1}{{(x - 1)}^{4}} = \frac{A}{x - 1} + \frac{B}{{(x - 1)}^{2}} + \frac{C}{{(x - 1)}^{3}} + \frac{D}{{(x - 1)}^{4}} then A + B + C =$

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Detailed Solution

$\begin{array}{rcl} \frac{3 x^{2} + x + 1}{{(x - 1)}^{4}} & = & \frac{A}{x - 1} + \frac{B}{{(x - 1)}^{2}} + \frac{C}{{(x - 1)}^{3}} + \frac{D}{{(x - 1)}^{4}} \\ x - 1 & = & y \Rightarrow x = y + 1 \\ \frac{3 {(y + 1)}^{2} + y + 1 + 1}{y^{4}} & = & \frac{3 y^{2} + 7 y + 5}{y^{4}} \\ = & \frac{3}{y^{2}} + \frac{7}{y^{3}} + \frac{5}{y^{4}} \\ = & \frac{3}{{(x - 1)}^{2}} + \frac{7}{{(x - 1)}^{3}} + \frac{5}{{(x - 1)}^{4}} \\ A & = & 0, B = 3, C = 7, D = 5 \\ A + B - C + D & = & 0 + 3 - 7 + 5 = 1 \end{array}$