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If (e^(x))/(1-x) = B(0) +B(1)x+B(2)x^(2)...

If `(e^(x))/(1-x) = B_(0) +B_(1)x+B_(2)x^(2)+...+B_(n)x^(n)+... `, then the value of `B_(n) - B_(n-1)` is

A

1

B

`1/n`

C

`1/(n!)`

D

None of these

Text Solution

Verified by Experts

The correct Answer is:
c
We have , ` e^(x) = (1-x) (B_(0) +B_(1)x+B_(2)x^(2)+...+`
` B_(n-1)x^(n-1) +B_(n)x^(n) +...)`
By the expression of `e^(x)`, we get
` 1+x/(1!)+(x^(2))/(2!) +...+(x^(n))/(n!)+...`
` = (1- x)(B_(0)+B_(1)x+B_(2)x^(2)+...+B_(n-1)x^(n-1) +B_(n)x^(n)+...)`
On equating the coefficient of `x^(n)` both sides , we get
` B_(n) - B_(n-1) = 1/(n!)`
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