Question

Find the coefficients of ${x^k}$ in the expansion of $\dfrac{{1 - 2x - {x^2}}}{{{e^{ - x}}}}$ is
A) $\dfrac{{1 - k - {k^2}}}{{k!}}$
B) $\dfrac{{{k^2} + 1}}{{k!}}$
C) \[\dfrac{1}{{k!}}\]
D) $\dfrac{{1 - k}}{{k!}}$

Answer 1

Hint: Expansion of a product of sums expressed as a sum of products by using the fact that multiplication distributes over the addition. First we calculate the expansion and then find the required answer. We know the expansion of the given term and we put this in the given data and we get the required answer.

Complete step by step answer:
First the collect the given data $\dfrac{{1 - 2x - {x^2}}}{{{e^{ - x}}}}$
$ = (1 - 2x - {x^2}){e^x}$
Now we find the expansion of ${e^x}$
\[\therefore {e^x} = 1 + \dfrac{x}{{1!}} + \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^3}}}{{3!}} + ..... + \dfrac{{{x^{k - 2}}}}{{(k - 2)!}} + \dfrac{{{x^{k - 1}}}}{{(k - 1)!}} + \dfrac{{{x^k}}}{{k!}} + ......\]
Therefore we multiply with the given data and we get
$(1 - 2x - {x^2})\left( {1 + \dfrac{x}{{1!}} + \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^3}}}{{3!}} + ..... + \dfrac{{{x^{k - 2}}}}{{(k - 2)!}} + \dfrac{{{x^{k - 1}}}}{{(k - 1)!}} + \dfrac{{{x^k}}}{{k!}} + ......} \right)$
Now multiply the above polynomial and find the coefficient of ${x^k}$ th term
$\therefore $ The coefficient of ${x^k}$ is $\dfrac{1}{{k!}} - \dfrac{2}{{(k - 1)!}} - \dfrac{1}{{(k - 2)!}}$
Now calculate this coefficient term and we get
$ = \dfrac{1}{{k!}} - \dfrac{2}{{(k - 1)!}} - \dfrac{1}{{(k - 2)!}}$
Multiplying $k$ in numerator and denominator of the second term and $k(k - 1)$ in numerator and denominator of the third term , we get
$ = \dfrac{1}{{k!}} - \dfrac{{2 \times k}}{{k \times (k - 1)!}} - \dfrac{{1 \times k \times (k - 1)}}{{k \times (k - 1) \times (k - 2)!}}$
\[ = \dfrac{1}{{k!}} - \dfrac{{2k}}{{k!}} - \dfrac{{{k^2} - k}}{{k!}}\]
We take the $lcm\left( {k!,k!,k!} \right) = k!$ and calculate we get
$ = \dfrac{{1 - 2k - {k^2} + k}}{{k!}}$
$ = \dfrac{{1 - k - {k^2}}}{{k!}}$
Therefore, the coefficients of ${x^k}$ in the expansion of $\dfrac{{1 - 2x - {x^2}}}{{{e^{ - x}}}}=\dfrac{{1 - k - {k^2}}}{{k!}}$. So, option (A) is correct.

Note:
We take care when we multiply these , we know ${x^a} \times {x^b} = {x^{a + b}}$ , therefore we take only those coefficients whose functions given ${x^k}$ . i.e., we take the $\dfrac{{{x^{k - 2}}}}{{(k - 2)!}}$ because ${x^{k - 2}} \times {x^2} = {x^{k - 2 + 2}} = {x^k}$ , we take $\dfrac{{{x^{k - 1}}}}{{(k - 1)!}}$ because ${x^{k - 1}} \times x = {x^{k - 1 + 1}} = {x^k}$ and we take $\dfrac{{{x^k}}}{{k!}}$ and take sum of them .