Question

If $y=\frac{x+1}{(x-1)(x-2)(x-3)}$, then $y_n=$

• A. $(-1{)}^{n}n![\frac{1}{(x-1{)}^{n+1}}-\frac{3}{(x-2{)}^{n+1}}+\frac{2}{(x-3{)}^{n+1}}]$
• B. $(-1{)}^{n}n![\frac{1}{(x-1{)}^{n+1}}+\frac{1}{(x-2{)}^{n+1}}+\frac{1}{(x+3{)}^{n+1}}]$
• C. $(-1{)}^{n}n![\frac{1}{(x-1{)}^n}-\frac{1}{(x-2{)}^n}+\frac{1}{(x-3{)}^n}]$
• D. $(-1{)}^n[\frac{1}{(x-1{)}^{n+1}}+\frac{1}{(x-2{)}^{n+1}}+\frac{1}{(x+3{)}^{n+1}}]$

Answer 1

The correct option is A $(-1{)}^{n}n![\frac{1}{(x-1{)}^{n+1}}-\frac{3}{(x-2{)}^{n+1}}+\frac{2}{(x-3{)}^{n+1}}]$

$\frac{2(x-1)-(x-2)+(x-2)-(x-3)}{(x-1)(x-2)(x-3)}$

$\frac{2}{(x-2)(x-3)}-\frac{1}{(x-1)(x-2)}$

$\frac{2\left(\right(x-2\left)\right(x-3\left)\right)}{(x-2)(x-3)}-\frac{(x-1)(x-2)}{(x-1)(x-2)}$

$\frac{2}{(x-3)}-\frac{2}{(x-2)}-\frac{1}{(x-2)}+\frac{1}{(x-1)}$

$y=\frac{-2}{(x-3)}-\frac{3}{(x-2)}+\frac{1}{(x-1)}$

$y^1=-\frac{2}{(x-3{)}^2}+\frac{3}{(x-2{)}^2}+\frac{1}{(x-1{)}^2}$

$y^{11}=+\frac{2!2}{(x-3{)}^3}+\frac{2!3}{(x-2{)}^3}+\frac{2!}{(x-1{)}^3}$

$y^{111}=-\frac{3!2}{(x-3{)}^4}+\frac{3!3}{(x-2{)}^4}+\frac{3!}{(x-1{)}^4}$

$y^n=(-1{)}^{n}n!(\frac{2}{(x-3{)}^{n+1}}-\frac{3}{(x-2{)}^{n+1}}+\frac{1}{(x-1{)}^{n+1}})$