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EAMCET Engineering 2007 Solved Paper

Section: Mathematics

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Question : 92 of 160

Marks: +1, -0

The coefficient of

x^k

in the expansion of

\frac{1-2x-x^2}{e^{-x}}

$\frac{1-k-k^2}{k!}$
$\frac{k^2+1}{k!}$
$\frac{1-k}{k!}$
$\frac{1}{k!}$

Solution:

Now,

\frac{1-2x-x^2}{e^{-x}}

= (1-2x-x^2)(e^x)

= (1-2x-x^2)\left(1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\dots\right)

+\frac{x^k}{k!}+\dots\infty)

= \left(1+x+\frac{x^2}{2!}+\dots+\frac{x^k}{k!}+\dots\infty\right) -2\left(x+x^2+\frac{x^3}{2!}\right.

+\dots+\frac{x^k}{(k-1)!}+\frac{x^{k+1}}{k!}+\dots\infty)

- \left(x^2+x^3+\frac{x^4}{2!}+\dots\right).

.

\frac{x^k}{(k-2)!}+\frac{x^{k+1}}{(k-1)!}+\frac{x^{k+2}}{k!}+\dots\infty)

\therefore

Coefficient of

x^k

in

\left(\frac{1-2x-x^2}{e^{-x}}\right)

=\frac{1}{k!}-\frac{2}{(k-1)!}-\frac{1}{(k-2)!}

=\frac{1}{k!}-\frac{2k}{k(k-1)!}-\frac{k(k-1)}{k(k-1)(k-2)!}

=\frac{1}{k!}-\frac{2k}{k!}-\frac{k(k-1)}{k!}

=\frac{1-k-k^2}{k!}

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