(1+x) (1+2x) (1+3x)(1-x) (1-2x) (1-3x)=k+A1-x+B1-2x+C1-3x k=6-6=-1 (1+x) (1+2x) (1+3x)=-(1-x) (1-2x) (1-3x)+A(1-2x) (1-3x)+B(1-x) (1-3x)+C(1-x) (1-2x) x=1 ⇒ 2(3) 4=A(-1) (-2) ⇒ A=12 x=12 ⇒ 32 (2) 52=B12 -12 ⇒ B=-30 x=13 ⇒ 43 5322=C23 13 ⇒ C=20 k+c=-1+20=19

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$If \frac{(1 + x) (1 + 2 x) (1 + 3 x)}{(1 - x) (1 - 5 x + 6 x^{2})} = k + \frac{A}{1 - x} + \frac{B}{1 - 2 x} + \frac{C}{1 - 3 x} then$

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$k + A + B + C = 0$

$k + A = 13$

$k + C = 19$

$k + B = 29$

answer is B.

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Detailed Solution

$\begin{array}{lcl} \frac{(1 + x) (1 + 2 x) (1 + 3 x)}{(1 - x) (1 - 2 x) (1 - 3 x)} \begin{array}{cl} = & k + \frac{A}{1 - x} + \frac{B}{1 - 2 x} + \frac{C}{1 - 3 x} \end{array} \\ \begin{array}{rcl} k & = & \frac{6}{- 6} = - 1 \end{array} \\ \begin{array}{rcl} (1 + x) (1 + 2 x) (1 + 3 x) & = & - (1 - x) (1 - 2 x) (1 - 3 x) + A (1 - 2 x) (1 - 3 x) + B (1 - x) (1 - 3 x) + C (1 - x) (1 - 2 x) \end{array} \\ \begin{array}{rcl} x & = & 1 \Rightarrow 2 (3) 4 = A (- 1) (- 2) \Rightarrow A = 12 \end{array} \\ x \begin{array}{cl} = & \frac{1}{2} \Rightarrow \frac{3}{2} (2) \frac{5}{2} = B (\frac{1}{2}) (\frac{- 1}{2}) \Rightarrow B = - 30 \end{array} \\ x \begin{array}{cl} = & \frac{1}{3} \Rightarrow \frac{4}{3} (\frac{5}{32}) 2 = C (\frac{2}{3}) (\frac{1}{3}) \Rightarrow C = 20 \end{array} \\ k + c \begin{array}{cl} = & - 1 + 20 = 19 \end{array} \end{array}$