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, $(A_1+A_3+A_5,A_2+A_4+A_6)$
$=(0+0-8,0+3+9)=(-8,12)$