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

 

Option: 1

(-1)^n n !\left[16(x-2)^{n-1}-(x-1)^{-n-1}\right]


Option: 2

(-1)^n n !\left[16(x-2)^{-n-1}+(x-1)^{-n-1}\right]


Option: 3

n !\left[16(x-2)^{-n-1}+(x-1)^{-n-1}\right]


Option: 4

None of these

 


Answers (1)

best_answer

y=\frac{x^4}{x^2-3 x+2}=x^2+3 x+7+\frac{15 x-14}{(x-1)(x-2)}=x^2+3 x+7-\frac{1}{(x-1)}+\frac{16}{(x-2)}

\begin{aligned} & \therefore y_n=D^n\left(x^2\right)+D^n(3 x)+D^n(7)-D^n\left[(x-1)^{-1}\right]+16 D^n\left[(x-2)^{-1}\right] \\ & =(-1)^n n !\left[-(x-1)^{-n-1}+16(x-2)^{-n-1}\right]=(-1)^n n !\left[16(x-2)^{-n-1}-(x-1)^{-n-1}\right] \end{aligned}

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shivangi.bhatnagar

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