∏ i = 0 2 5 ( x + i ) 2 5 ! = i = 0 ∑ 2 5 x + i A i
The above shows a partial fractions decomposition for constants A 0 , A 1 , … , A 2 5 .
Find A 2 4 .
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We note that the RHS is the partial fractions of the LHS.
∏ i = 0 2 5 ( x + i ) 2 5 ! 2 5 ! 2 5 ! ⟹ A 2 4 = i = 0 ∑ 2 5 x + i A i = i = 0 ∑ 2 5 A i k = i ∏ 2 5 ( x + k ) Putting x = − 2 4 , all terms in the RHS with ( x + 2 4 ) become 0 except 24th term. = A 2 4 ( − 2 4 ) ( − 2 3 ) ( − 2 2 ) . . . ( − 2 ) ( − 1 ) ( 1 ) = 2 4 ! A 2 4 = 2 5