Let fx - a0-a1x-a2x2-anxn-and fx-1-x - b0-b1x-b2x2-bnxn- then

The correct option is D none of these f(x)=a_0+a_1x+a_2x^2.. f(x)(1 x)^ 1=(1+x+x^2..∞)(a_0+a_1x+a_2x^2...) =(a_0+(a_0+a_1)x+(a_0+a_1+a_2)x^2...) C ...

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Byju's Answer

Standard XII

Mathematics

Integration by Partial Fractions

Let fx = a0...

Question

Let $f (x) = a_{0} + a_{1} x + a_{2} x^{2} + . . . + a_{n} x^{n} = . . .$ and $\frac{f (x)}{1 - x} = b_{0} + b_{1} x + b_{2} x^{2} + . . . + b_{n} x^{n} + . . .$ , then

b_{n} + b_{n - 1} = a_{n}

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b_{n} - b_{n - 1} = a_{n + 1}

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\frac{b_{n}}{b_{n - 1}} = a_{n}

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none of these

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Solution

The correct option is D none of these
$f (x) = a_{0} + a_{1} x + a_{2} x^{2} . .$
$f (x) (1 - x)^{- 1} = (1 + x + x^{2} . . \infty) (a_{0} + a_{1} x + a_{2} x^{2} . . .)$
$= (a_{0} + (a_{0} + a_{1}) x + (a_{0} + a_{1} + a_{2}) x^{2} . . .)$
Comparing coefficients, we get
$b_{0} = a_{0}$
$b_{1} = a_{0} + a_{1}$
$b_{2} = a_{0} + a_{1} + a_{2}$
:
:
Hence $b_{n + 1} - b_{n} = a_{n + 1}$

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Similar questions

Q. If

\frac{e^{x}}{1 - x} = B_{0} + B_{1} x + B_{2} x^{2} + . . . + B_{n - 1} x^{n - 1} + B_{n} x^{n} + . . .

Then

B_{n} - B_{n - 1} = \frac{b}{n!}

. Find

b

Q. If

\frac{e^{x}}{1 - x} = B_{0} + B_{1} x + B_{2} x^{2} + . . . . + B_{n} x^{n} + . . . .,

then

B_{n} - B_{n - 1}

equals

Q. Let

f (x) = a_{0} + a_{1} x + a_{2} x^{2} + . . . . . + a_{n} x^{n} + . . . . . . .

and

\frac{f (x)}{1 - x} = b_{0} + b_{1} x + b_{2} x^{2} + . . . . . . . + b_{n} x^{n} + . . . . .

then

Q. If

\frac{e^{x}}{a - x} = B_{0} + B_{1} x + B_{2} x^{2} + . . . + B_{n} x^{n} + . . .

, find the value of

B_{n} - B_{n - 1}

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a

Q. Consider two polynomials

P (x) = a_{n} x^{n} + a_{n - 1} x^{n - 1} + \dots + a_{1} x + a_{0}

, and

Q (x) = b_{n} x^{n} + b_{n - 1} x^{n - 1} + \dots + b_{1} x + b_{0}

with integer coefficients such that

a_{n} - b_{n}

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a_{n - 1} = b_{n - 1}

and

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Let f(x)=a0+a1x+a2x2+...+anxn=...and f(x)1−x=b0+b1x+b2x2+...+bnxn+..., then

The correct option is D none of thesef(x)=a0+a1x+a2x2..f(x)(1−x)−1=(1+x+x2..∞)(a0+a1x+a2x2...)=(a0+(a0+a1)x+(a0+a1+a2)x2...)Comparing coefficients, we getb0=a0b1=a0+a1b2=a0+a1+a2::Hence bn+1−bn=an+1

Let $f (x) = a_{0} + a_{1} x + a_{2} x^{2} + . . . + a_{n} x^{n} = . . .$ and $\frac{f (x)}{1 - x} = b_{0} + b_{1} x + b_{2} x^{2} + . . . + b_{n} x^{n} + . . .$ , then