\prod _{ }^{ }{ } & \sum { }

Algebra Level pending

25 ! i = 0 25 ( x + i ) = i = 0 25 A i x + i \large \dfrac {25!}{\prod_{i=0}^{25}(x+i)} = \sum_{i=0}^{25} \frac {A_i}{x+i}

The above shows a partial fractions decomposition for constants A 0 , A 1 , , A 25 A_0, A_1, \ldots , A_{25} .

Find A 24 A_{24} .


The answer is 25.

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1 solution

We note that the RHS is the partial fractions of the LHS.

25 ! i = 0 25 ( x + i ) = i = 0 25 A i x + i 25 ! = i = 0 25 A i k i 25 ( x + k ) Putting x = 24 , all terms in the RHS with ( x + 24 ) become 0 except 24th term. 25 ! = A 24 ( 24 ) ( 23 ) ( 22 ) . . . ( 2 ) ( 1 ) ( 1 ) = 24 ! A 24 A 24 = 25 \begin{aligned} \frac {25!}{\prod_{i=0}^{25} (x+i)} & = \sum_{i=0}^{25} \frac {A_i}{x+i} \\ 25! & = \sum_{i=0}^{25} A_i \prod_{k \ne i}^{25} (x+k) \quad \quad \small {\color{#3D99F6}\text{Putting }x = -24 \text{, all terms in the RHS with }(x+24) \text{ become }0 \text{ except 24th term.}} \\ 25! & = A_{24}(-24)(-23)(-22)...(-2)(-1)(1) \\ & = 24!A_{24} \\ \implies A_{24} & = \boxed{25} \end{aligned}

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