If x - 1- then the coefficient of xn in the expansion of ex-1 -x is 1 - a-1- - a2-2- - - - an-2-n-2- - an-1-n-1- -Find a-7

We have, #160;e^x/1 x= #160;e^x(1 x)^ 1 #160; #160; #160; #160; #160; #160; #160; =( 1 + x/1! + x^2/2! + ... + x^n 2/(n 2)! + x ...

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Byju's Answer

Standard XII

Mathematics

Summation by Sigma Method

If x < 1, t...

Question

If $x < 1$ , then the coefficient of $x^{n}$ in the expansion of $\frac{e^{x}}{1 - x}$ is $1 + \frac{a}{1!} + \frac{a^{2}}{2!} + . . . + \frac{a^{n - 2}}{(n - 2)!} + \frac{a^{n - 1}}{(n - 1)!}$ .
Find $a + 7$

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Solution

We have,
$\frac{e^{x}}{1 - x} = e^{x} (1 - x)^{- 1}$
$= (1 + \frac{x}{1!} + \frac{x^{2}}{2!} + . . . + \frac{x^{n - 2}}{(n - 2)!} + \frac{x^{n - 1}}{(n - 1)!} + \frac{x^{n}}{n!} + . . . \infty) \times (1 + x + x^{2} + . . . + x^{n - 2} + x^{n - 1} + x^{n} + . . . \infty)$

Coefficient of $x^{n}$ in $\frac{e^{x}}{1 - x}$
$= 1 + \frac{1}{1!} + \frac{1}{2!} + . . . + \frac{1}{(n - 1)!} + \frac{1}{n!}$
$∴ a = 1$

$\Rightarrow a + 7 = 1 + 7 = 8$

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If x<1, then the coefficient of xn in the expansion of ex1−x is 1+a1!+a22!+...+an−2(n−2)!+an−1(n−1)! .Find a+7

We have,ex1−x=ex(1−x)−1 =(1+x1!+x22!+...+xn−2(n−2)!+xn−1(n−1)!+xnn!+...∞)×(1+x+x2+...+xn−2+xn−1+xn+...∞)Coefficient of xn in ex1−x =1+11!+12!+...+1(n−1)!+1n!∴a=1⇒a+7=1+7=8

If $x < 1$ , then the coefficient of $x^{n}$ in the expansion of $\frac{e^{x}}{1 - x}$ is $1 + \frac{a}{1!} + \frac{a^{2}}{2!} + . . . + \frac{a^{n - 2}}{(n - 2)!} + \frac{a^{n - 1}}{(n - 1)!}$ .
Find $a + 7$