S = 3 ! 8 − 5 ! 3 2 + 7 ! 1 2 8 − 9 ! 5 1 2 + ⋯
If S as defined above can be represented as α − sin β for some positive integers α and β , then find α + 1 9 β − 1 .
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instead,why dont u simply consider the expansion for s i n x ??
Relevant wiki: Maclaurin Series
S = 3 ! 8 − 5 ! 3 2 + 7 ! 1 2 8 − 9 ! 5 1 2 + ⋯ = 3 ! 2 3 − 5 ! 2 5 + 7 ! 2 7 − 9 ! 2 9 + ⋯ = 2 − ( 1 ! 2 − 3 ! 2 3 + 5 ! 2 5 − 7 ! 2 7 + ⋯ ) = 2 − sin 2 Maclaurin’s series
⟹ α + 1 9 β − 1 = 2 + 1 9 ( 2 ) − 1 = 3 9
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Consider the expansion,
cos x = 1 − 2 ! x 2 + 4 ! x 4 − 6 ! x 6 + . . . . .
cos x − 1 = − 2 ! x 2 + 4 ! x 4 − 6 ! x 6 + . . . . .
1 − cos x = 2 ! x 2 − 4 ! x 4 + 6 ! x 6 − . . . . .
Integrating on both the sides from 0 to 2,
∫ 0 2 ( 1 − cos x ) d x = ∫ 0 2 ( 2 ! x 2 − 4 ! x 4 + 6 ! x 6 − . . . . . ) d x
2 − sin 2 = 3 . 2 ! 2 3 − 5 . 4 ! 2 5 + 7 . 6 ! 2 7 − . . . . .
∴ 2 − sin 2 = 3 ! 8 − 5 ! 3 2 + 7 ! 1 2 8 − . . . . .
Hence, α = 2 , β = 2