If 1/x(x+1)(x+2)·s(x+n) = A₀/x + A₁/x+1 + A₂/x+2 + ·s + Aₙ/x+n then A_r is equal to

Correct Answer is : r!(1)r(n−r)!r! (1)^r/(n-r)!(n−r)!r!(1)r

Correct Answer is : r!(1)r(n−r)!r! (1)^r/(n-r)!(n−r)!r!(1)r

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Manipal Entrance Test 2013 Math Paper

Question : 67 of 80

Marks: +1, -0

\frac{1}{x(x+1)(x+2)\cdots(x+n)}

= \frac{A_0}{x} + \frac{A_1}{x+1}

+ \frac{A_2}{x+2} + \cdots + \frac{A_n}{x+n}

A_r

is equal to

Solution:

We have

\frac{1}{x(x+1)(x+2)\cdots(x+n)}

= \frac{A_0}{x} + \frac{A_1}{x+1}

+ \cdots + \frac{A_r}{x+r} + \cdots + \frac{A_n}{x+n}

⇒

1 = A_0 (x+1)(x+2)\cdots(x+n)

+ \cdots + A_r x(x+1)(x+2)\cdots(x+r-1)(x+r+1)

\cdots(x+n) + \cdots + A_n x(x+1)\cdots(x+n-1)

Putting

x = -r,

we get:

1 = A_r (-r)(-r+1)(-r+2)\cdots

(-3)(-2)(-1)(1)(2)

\cdots (n-r)

⇒

1 = A_r (-1)^r \{ r(r-1)(r-2) \cdots (3\cdot2\cdot1) \}

(1\cdot2\cdot3\cdots (n-r))

⇒

1 = A_r (-1)^r \{ r(r-1)(r-2) \cdots (3\cdot2\cdot1) \}

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