Question

If $\frac{3x^{2} -2x +4}{ \left(x+1\right)^{6}} = \frac{A_{1}}{x +1} + \frac{A_{2}}{\left(x+1\right)^{2}} + \frac{A_{3}}{\left(x+1\right)^{3}} + \frac{A_{4}}{\left(x+1\right)^{4}} + \frac{A_{5}}{\left(x+1\right)^{5}} + \frac{A_{6}}{\left(x+1\right)^{6}}$ , then $ \left(A_{1} + A_{3} +A_{5} , A_{2} + A_{4} + A_{6}\right) = $

• A. (0, 0)
• B. (-8, -12)
• C. (8, -12)
• D. (-8, 12)

Answer 1

Answer: (D)

$\frac{3x^{2} -2x +4}{ \left(x+1\right)^{6}} = \frac{A_{1}}{x +1} + \frac{A_{2}}{\left(x+1\right)^{2}} + \frac{A_{3}}{\left(x+1\right)^{3}} + \frac{A_{4}}{\left(x+1\right)^{4}} + \frac{A_{5}}{\left(x+1\right)^{5}} + \frac{A_{6}}{\left(x+1\right)^{6}}$ $3x^{2} -2x +4 = A_{1} \left(x+1\right)^{5} +A_{2} \left(x +1\right)^{4} +A_{3}\left(x+1\right)^{3} +A_{4}\left(x+1\right)^{2}+A_{5}\left(x+1\right)+A_{6} \, \, \, $ ....(i)
Putting $x = - 1$, we get
$ 3\left(-1\right)^{2} -2\left(-1\right)+ 4= A_{6} \Rightarrow A_{6} =9 $
Putting the value of $A_6$ in (i), we get
$3x^2 - 2x + 4 - 9 = (x + 1) [A_1(x + 1)^4 + A_2 (x + 1)^3 + A_3 (x + 1)^2 + A_4 (x + 1) + A_5]$
$3x^2 - 2x - 5 = (x + 1) [A_1(x + 1)^4 + A_2 (x + 1)^3 + A_3 (x + 1)^2 + A_4 (x + 1) + A_5]$ $3x^2 - 5x + 3x - 5 = (x + 1) [A_1(x + 1)^4 + A_2 (x + 1)^3 + A_3 (x + 1)^2 + A_4 (x + 1) + A_5]$
$(3x -5) (x + 1) = (x + 1)[A_1(x + 1)^4 + A_2 (x + 1)^3 + A_3 (x + 1)^2 + A_4 (x + 1) + A_5]$
$3x- 5 = A_1(x + 1)4 + A_2 (x + 1)^3 + A_3 (x + 1)^2 + A_4 (x + 1) + A_5] \, \, \, \, \, \, $ ...(ii)
Putting $x = -1$ in (ii), we get $\Rightarrow \, A_5= -8$
Putting the value of $A_5$ in (ii), we get
$3x - 5 + 8 = (x + 1) [A_1 (x + 1)^3 + A_2 (x + 1)^2 + A_3 (x + 1) + A_4]$
$3 = A_1( x + 1)^3 + A_2(x + 1)^2 + A_3 (x + 1) + A_4$ .. (iii)
Putting $x = -1$ in (iii), we get $A_4 = 3$
Putting the value of $A_4$ in (iii), we get
$0 = (x + l)[A_1(x + 1)^2 + A_2 (x + 1) + A_3]$
$\Rightarrow \, \, A_3 = 0$
Similarly, $A_1 = A_2 = 0$
Now, $(A_1 + A_3 + A_5,A_2 +A_4 +A_6)$
$ = (0 +0 -8, 0+ 3 +9) =(-8 ,12)$