What series is this?

Calculus Level 4

S = 8 3 ! 32 5 ! + 128 7 ! 512 9 ! + \large S=\frac 8{3!} - \frac{32}{5!} + \frac{128}{7!} - \frac{512}{9!} + \cdots

If S S as defined above can be represented as α sin β \alpha - \sin \beta for some positive integers α \alpha and β \beta , then find α + 19 β 1 \alpha + 19 \beta -1 .


The answer is 39.

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2 solutions

Sparsh Sarode
Dec 19, 2016

Consider the expansion,

cos x = 1 x 2 2 ! + x 4 4 ! x 6 6 ! + . . . . . \cos x =1-\dfrac{x^2}{2!}+\dfrac{x^4}{4!}-\dfrac{x^6}{6!}+.....

cos x 1 = x 2 2 ! + x 4 4 ! x 6 6 ! + . . . . . \cos x -1=-\dfrac{x^2}{2!}+\dfrac{x^4}{4!}-\dfrac{x^6}{6!}+.....

1 cos x = x 2 2 ! x 4 4 ! + x 6 6 ! . . . . . 1- \cos x =\dfrac{x^2}{2!}-\dfrac{x^4}{4!}+\dfrac{x^6}{6!}-.....

Integrating on both the sides from 0 to 2,

0 2 ( 1 cos x ) d x = 0 2 ( x 2 2 ! x 4 4 ! + x 6 6 ! . . . . . ) d x \displaystyle \int_0 ^2 (1- \cos x)dx =\int_0 ^2 \Big ( \dfrac{x^2}{2!}-\dfrac{x^4}{4!}+\dfrac{x^6}{6!}-..... \Big ) dx

2 sin 2 = 2 3 3.2 ! 2 5 5.4 ! + 2 7 7.6 ! . . . . . 2 - \sin 2 = \dfrac{2^3}{3.2!}-\dfrac{2^5}{5.4!}+\dfrac{2^7}{7.6!}-.....

2 sin 2 = 8 3 ! 32 5 ! + 128 7 ! . . . . . \therefore 2- \sin 2 = \dfrac{8}{3!}-\dfrac{32}{5!}+\dfrac{128}{7!}-.....

Hence, α = 2 , β = 2 \color{#3D99F6}{ \boxed{ \alpha =2, \beta=2}}

instead,why dont u simply consider the expansion for s i n x sinx ??

Rohith M.Athreya - 4 years, 5 months ago

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Oh! Yea.. Thts a direct one...

Sparsh Sarode - 4 years, 5 months ago
Chew-Seong Cheong
Dec 19, 2016

Relevant wiki: Maclaurin Series

S = 8 3 ! 32 5 ! + 128 7 ! 512 9 ! + = 2 3 3 ! 2 5 5 ! + 2 7 7 ! 2 9 9 ! + = 2 ( 2 1 ! 2 3 3 ! + 2 5 5 ! 2 7 7 ! + ) Maclaurin’s series = 2 sin 2 \begin{aligned} S & = \frac 8{3!} - \frac {32}{5!} + \frac {128}{7!} - \frac {512}{9!} + \cdots \\ & = \frac {2^3}{3!} - \frac {2^5}{5!} + \frac {2^7}{7!} - \frac {2^9}{9!} + \cdots \\ & = 2 - \color{#3D99F6} \left(\frac {2}{1!} - \frac {2^3}{3!} + \frac {2^5}{5!} - \frac {2^7}{7!} + \cdots \right) & \small \color{#3D99F6} \text{Maclaurin's series} \\ & = 2 - \color{#3D99F6} \sin 2 \end{aligned}

α + 19 β 1 = 2 + 19 ( 2 ) 1 = 39 \implies \alpha + 19 \beta - 1 = 2+19(2) - 1= \boxed{39}

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