In the expansion of a - bxex- the coefficient of xr is

The correct option is C ( 1)^ra brr! (a + bx)e^ x = (a + bx) #160; {1 x1! + x^22! x^33! + ... + ( 1)^nx^nn! + .... } ∴ The co ...

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Standard XII

Mathematics

Binomial Theorem for Any Index

In the expans...

Question

In the expansion of $\frac{a + b x}{e^{x}}$ , the coefficient of $x^{r}$ is

\frac{a - b}{r!}

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\frac{a - b r}{r!}

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(- 1)^{r} \frac{a - b r}{r!}

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None of these

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Solution

The correct option is C $(- 1)^{r} \frac{a - b r}{r!}$
$(a + b x) e^{- x}$

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

$∴$ The coefficient of $x^{r} =$ $a \frac{(- 1)^{r}}{r!} + b \frac{(- 1)^{r - 1}}{(r - 1)!}$
$= a \frac{(- 1)^{r}}{r!} - b r \frac{(- 1)^{r}}{(r)!}$
$= \frac{(- 1)^{r}}{r!} (a - b r)$

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In the expansion of a+bxex, the coefficient of xr is

The correct option is C (−1)ra−brr!(a+bx)e−x=(a+bx) {1−x1!+x22!−x33!+...+(−1)nxnn!+....}∴ The coefficient of xr= a(−1)rr!+b(−1)r−1(r−1)! =a(−1)rr!−br(−1)r(r)! =(−1)rr!(a−br)

In the expansion of $\frac{a + b x}{e^{x}}$ , the coefficient of $x^{r}$ is