Question

If $y = \frac{x^{4}}{x^{2} - 3x + 2}$, then for n > 2 the value of y_n is equal to

• A. $( - 1)^{n}n!\lbrack 16(x - 2)^{- n - 1} - (x - 1)^{- n - 1}\rbrack$
• B. $( - 1)^{n}n!\lbrack 16(x - 2)^{- n - 1} + (x - 1)^{- n - 1}\rbrack$
• C. $n!\lbrack 16(x - 2)^{- n - 1} + (x - 1)^{- n - 1}\rbrack$
• D. None of these

Answer 1

Answer: (A)

$( - 1)^{n}n!\lbrack 16(x - 2)^{- n - 1} - (x - 1)^{- n - 1}\rbrack$

Answer 2

$y = \frac{x^{4}}{x^{2} - 3x + 2} = x^{2} + 3x + 7 + \frac{15x - 14}{(x - 1)(x - 2)}$

= $x^{2} + 3x + 7 - \frac{1}{(x - 1)} + \frac{16}{(x - 2)}$

∴$y_{n} = D^{n}(x^{2}) + D^{n}(3x) + D^{n}(7) - D^{n}\lbrack(x - 1)^{- 1}\rbrack + 16D^{n}\lbrack(x - 2)^{- 1}\rbrack = ( - 1)^{n}n!\lbrack - (x - 1)^{- n - 1} + 16(x - 2)^{- n - 1}\rbrack$

= $( - 1)^{n}n!\lbrack 16(x - 2)^{- n - 1} - (x - 1)^{- n - 1}\rbrack$.