The coefficient of xn in the series 1 - a-bx1- - a-bx22- - a-bx33- - -

The correct option is A e^a . b^nn! As we know that, e^x=1+x1!+x^22!+x^33!+..... ∴ 1 + a+bx1! + (a+bx)^22! + (a+bx)^33! ... ∞ = e^a+bx Now, coef ...

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Standard XII

Mathematics

Binomial Theorem for Any Index

The coefficie...

Question

The coefficient of $x^{n}$ in the series
$1 + \frac{a + b x}{1!} + \frac{(a + b x)^{2}}{2!} + \frac{(a + b x)^{3}}{3!} . . . \infty$

\frac{(a b)^{n}}{n!}

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e^{b} \frac{a^{n}}{n!}

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e^{a} . \frac{b^{n}}{n!}

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e^{a + b} . \frac{(a b)^{n}}{n!}

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Solution

The correct option is A $e^{a} . \frac{b^{n}}{n!}$
As we know that, $e^{x} = 1 + \frac{x}{1!} + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + . . . . .$

$∴ 1 + \frac{a + b x}{1!} + \frac{(a + b x)^{2}}{2!} + \frac{(a + b x)^{3}}{3!} . . . \infty = e^{a + b x}$
Now, coefficient of $x^{n}$ in $e^{a x}$ is $\frac{a^{n}}{n!}$
$∴$ Coefficient of $x^{n}$ in $e^{a} e^{b x} = e^{a} . \frac{(b)^{n}}{n!}$
Hence, option C is correct.

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The coefficient of xn in the series 1+a+bx1!+(a+bx)22!+(a+bx)33!...∞

The correct option is A ea.bnn!As we know that, ex=1+x1!+x22!+x33!+.....∴1+a+bx1!+(a+bx)22!+(a+bx)33!...∞=ea+bxNow, coefficient of xn in eax is ann!∴ Coefficient of xn in eaebx=ea.(b)nn!Hence, option C is correct.

The coefficient of $x^{n}$ in the series
$1 + \frac{a + b x}{1!} + \frac{(a + b x)^{2}}{2!} + \frac{(a + b x)^{3}}{3!} . . . \infty$