If x4x-1x-2x-3= Ax+B.1x-1+C1x-2+D.1x-3+E, then A+B+C+D+E=

x4(x-1) (x-2) (x-3)=Ax+E+Bx-1+Cx-2+Cx-1 Ax+E=x+6 A=1, E=6 x4=(x-16) (x-1) (x-2) (x-3)+B(x-2) (x-3)+C(x-1) (x-3)+D(x-1) (x-2) x=1 ⇒ 1=2B ⇒ B=12 x=2 ⇒ 16=-C ⇒ C=-16 x=3 ⇒ 81=2D ⇒ D=812 A+B+C+D+E=1+6+12-16+812 =48-16=32

If x4x-1x-2x-3= Ax+B.1x-1+C1x-2+D.1x-3+E, then A+B+C+D+E=

x4(x-1) (x-2) (x-3)=Ax+E+Bx-1+Cx-2+Cx-1 Ax+E=x+6 A=1, E=6 x4=(x-16) (x-1) (x-2) (x-3)+B(x-2) (x-3)+C(x-1) (x-3)+D(x-1) (x-2) x=1 ⇒ 1=2B ⇒ B=12 x=2 ⇒ 16=-C ⇒ C=-16 x=3 ⇒ 81=2D ⇒ D=812 A+B+C+D+E=1+6+12-16+812 =48-16=32

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

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Detailed Solution

$\frac{x^{4}}{(x - 1) (x - 2) (x - 3)} = Ax + E + \frac{B}{x - 1} + \frac{C}{x - 2} + \frac{C}{x - 1} Ax + E = x + 6 A = 1, E = 6 x^{4} = (x - 16) (x - 1) (x - 2) (x - 3) + B (x - 2) (x - 3) + C (x - 1) (x - 3) + D (x - 1) (x - 2) x = 1 \Rightarrow 1 = 2 B \Rightarrow B = \frac{1}{2} x = 2 \Rightarrow 16 = - C \Rightarrow C = - 16 x = 3 \Rightarrow 81 = 2 D \Rightarrow D = \frac{81}{2} A + B + C + D + E = 1 + 6 + \frac{1}{2} - 16 + \frac{81}{2} = 48 - 16 = 32$